Trigonometrical Equations and Inequations PDF

46 Trigonometrical Equations and 60 Inequations 2.1 Introduction. An equation involving one or more trigonometrical ratio of an unknown angle is called a E3   trigonometrical equation i.e., sin x  cos 2 x  1 , (1  tan  )(1  sin 2 )  1  tan  ; | sec   |  2 etc. 4  A trigonometric e...

46 Trigonometrical Equations and 60 Inequations 2.1 Introduction. An equation involving one or more trigonometrical ratio of an unknown angle is called a E3   trigonometrical equation i.e., sin x  cos 2 x  1 , (1  tan  )(1  sin 2 )  1  tan  ; | sec   |  2 etc. 4  A trigonometric equation is different from a trigonometrical identities. An identity is ID satisfied for every value of the unknown angle e.g., cos 2 x  1  sin 2 x is true x  R while a trigonometric equation is satisfied for some particular values of the unknown angle. (1) Roots of trigonometrical equation : The value of unknown angle (a variable quantity) which 1 satisfies the given equation is called the root of an equation e.g., cos   , the root is   60 o or 2 U   300 o because the equation is satisfied if we put   60 o or   300 o. D YG (2) Solution of trigonometrical equations : A value of the unknown angle which satisfies the trigonometrical equation is called its solution. Since all trigonometrical ratios are periodic in nature, generally a trigonometrical equation has more than one solution or an infinite number of solutions. There are basically three types of solutions: (i) Particular solution : A specific value of unknown angle satisfying the equation. U (ii) Principal solution : Smallest numerical value of the unknown angle satisfying the equation (Numerically smallest particular solution.) (iii) General solution : Complete set of values of the unknown angle satisfying the equation. It contains all particular solutions as well as principal solutions. ST When we have two numerically equal smallest unknown angles, preference is given to the 2 positive value in writing the principal solution. e.g., has sec   3   11 11 23  23  etc. , , , , , 6 6 6 6 6 6 As its particular solutions out of these, the numerically smallest are   6  6 and   6 Y but the principal solution is taken as P /6 to write the general solution we notice that the position on O X – /6 P Trigonometrical Equations and Inequations 47 P or P' can be obtained by rotation of OP or OP around O through a complete angle (2 ) any number of times and in any direction (clockwise or anticlockwise)  6 ,k  Z. 2.2 General Solution of Standard Trigonometrical Equations. 60  The general solution is   2k   E3 (1) General solution of the equation sin  = sin: If sin   sin  or sin   sin   0                 or, 2 sin   sin    0 or cos  0  cos  0  2   2   2   2       or,  m  ; m  I or  (2m  1) ; m  I 2 2 2   ( 2 2 m m  1 )   ; m ; mI  I        or Note ID   = (any even multiple of  ) +  or  = (any odd multiple of ) –  θ  nπ  (1)n α ; n  I :  The equation cosec   cosec  is equivalent to sin   sin . So these two equation having the same general solution. (2) General solution of the equation cos = cos : If cos   cos  D YG U  cos   cos   0                       2 sin   n ; n  I  n  ; n  I or . sin    0  sin  0,    0 or sin  2 2  2   2   2   2     2n    ; n  I or   2n    ; n  I. for the general solution of cos   cos  , combine these two result which gives θ  2nπ  α; n  I Note :  The equation sec   sec  is equivalent to cos   cos , so the general solution of these two equations are same. sin  sin   cos  cos  θ  nπ  α; n  I (3) General solution of the equation tan  = tan  : If tan   tan   U  sin  cos   cos  sin   0  sin(   )  0      n ; n  I Note :  The equation cot   cot  is equivalent to tan   tan  so these two equations having the same general solution. ST 2.3 General Solution of Some Particular Equations.   (1) sin   0    n  , cos   0    (2n  1) or n   , 2 (2) sin   1    (4 n  1)  (3) sin   1    (4 n  3) tan   1    (4 n  1)  4 or 2n  2  2 2 tan   0    n  , cos   1    2n , tan   1    (4 n  1) 2n   or or n  (4) tan  = not defined  2 3 , 2 4  2 , cot  = not defined 4 or n    4 cos   1    (2n  1) ,     (2n  1)     n 48 Trigonometrical Equations and Inequations cosec  = not defined    n  , sec  = not defined    (2n  1)  2. Important Tips  For equations involving two multiple angles, use multiple and sub-multiple angle formulas, if necessary. 3 , then the general value of  is 2 (a) 2n    (b) 2n   6  (c) n   (1)n 3 3    sin   sin    n   (1)n. 2 3 3 Solution: (c) sin   Example: 2 The general solution of tan 3 x  1 is (a) n    4 (b)   3 x  n   n   3 12 If sin 3  sin  , then the general value of 3 U  (b) n  , (2n  1)  (c) n  , (2n  1) 4 (d) n    4  3 (d) None of these 2n   n 2 And for (m) odd, i.e., m  (2n  1) then   (2n  1)  4. The general solution of 2 sin 2   3 sin  2  0 is  U (a) n   (1)n Solution: (d) 6 sin 3  sin  or 3  m   (1)m  For (m) even i.e., m  2n then   Example: 4   is D YG (a) 2n  , (2n  1) x  (d) n   (1)n n  . 3 12 Example: 3 4 3 (c) n  tan 3 x  tan 4  [Karnataka CET 1991] Solution: (b) Solution: (b) [MP PET 1988] E3 If sin   ID Example: 1 60  For equations involving more than two multiple angles (i) Apply C  D formula to combine the two.(ii) Choose such pairs of multiple angle so that after applying the above formulae we get a common factor in the equation. 2 (b) n   (1)n  6 [Roorkee 1993] (c) n   (1)n 7 6 (d) n   (1)n  6 2 sin 2   3 sin  2  0  2 sin 2   4 sin   sin   2  0  2 sin  (sin   2)  (sin   2)  0 ST (2 sin  1)(sin  2)  0 sin   2 Example: 5 (which is impossible) 1  sin  sin( / 6)    n  (1)n  / 6. 2 The number of solutions of the equation 3 sin 2 x  7 sin x  2  0 in the interval [0, 5 ] is [MP PET 2001; IIT 1998] (a) 0 Solution: (c)  sin    (b) 5 (c) 6 (d) 10 3 sin x  7 sin x  2  0  3 sin x  6 sin x  sin x  2  0  (3 sin x  1)(sin x  2)  0 , 2 But sin x  2 so sin x  2 1. Hence from 0 to 2  2 solution's (one in Ist quadrant and other in 2nd quadrant), 3 from 2 to 4  2 solution's and 4 to 5  2 solution's. So total number of solutions  6. Example: 6 Number of solutions of the equation tan x  sec x  2 cos x , lying in the interval 0, 2  is [AIEEE 2002; MP PET 2000; IIT 1993] Trigonometrical Equations and Inequations (a) 0 Solution: (c) (b) 1 (c) 2 49 (d) 3 sin x 1   2 cos x  sin x  1  2 cos 2 x  2 sin 2 x  sin x  1  0  2 sin x  1sin x  1  0 cos x cos x So, sin x  1 or sin x   5 3 3 1 or x  , but does not satisfy the equation, So total number  x  2 2 6 6 2 If tan   tan 2  3 tan  tan 2  3 , then (a)   (c)   (3n  1) , n  I (d) None of these 9  tan   tan 2  3 (1  tan  tan 2 ) E3 tan   tan 2 n       3  tan 3  tan    3  n       (3 n  1). 1  tan  tan 2 3 9 3 9 3 The solution of the equation cos 2 x  2 cos x  4 sin x  sin 2 x , (0  x   ) is 1 (a)   cot 1   2 Solution: (c) (6 n  1) , n  I 9 tan   tan 2  3 tan  tan 2  3  Example: 8 (b)   (b)   tan 1 (2) ID Solution: (c) (6 n  1) , n  I 18 [UPSEAT 2001]  1 (c)   tan 1     2 [DCE 2001] (d) None of these Given equation is cos 2 x  2 cos x  4 sin x  sin 2 x U Example: 7 60 of solutions  2.  cos 2 x  2 cos x  4 sin x  2 sin x cos x  cos x(cos x  2)  2 sin x(2  cos x )  (cos x  2)(cos x  2 sin x )  0  tan x   (cos x  2) D YG  cos x  2 sin x  0 1  1  x  n   tan 1    , n  I 2  2  1 As 0  x   , therefore x    tan 1   .  2 Example: 9 cos  The solution of the equation  sin   cos  U (a)   n  Example: 10 (c)   n   (1)n  2  4 (d)   2n    4.  6 (b) 2n    4 (c) n   (1)n  3 (d) n   (1)n  4 2 3 cos 2   sin   2 3 sin 2   sin   2 3  0  sin   1  7 4 3 Example: 11 2 The general value of  in the equation 2 3 cos   tan  , is (a) 2n   Solution: (c)  cos  sin   0, is cos  After solving the determinant 2 cos   0    2n   ST Solution: (b) (b)   2n   sin  cos   sin   sin   8 (impossible) or sin   4 3 6 4 3  3 2    n   (1)n The general value of  is obtained from the equation cos 2  sin  is (a) 2   2    (b)   2n       2   (c)   n   (1)n  2  3. [MP PET 1996]    (d)   n      4 2 50 Trigonometrical Equations and Inequations Solution: (d)   cos 2  sin   cos 2  cos     2        2  2n          n      2 4 2   Solution: (a)  cos 2 B sin B  0, then B  cos B [EAMCET 2003] (c) (2n  1) (b) n  2 On expanding the determinant cos 2 ( A  B)  sin 2 ( A  B)  cos 2 B  0 1  cos 2B  0 or cos 2 B  cos  or 2 B  2n   or B  (2n  1) 2. 2 3 (b)   2m    4 ID If cos   cos 2  cos 3  0, then the general value of  is (a)   2m   Solution: (a)  (c)   m   (1)m 2 3 (d)   m   (1)m  3  cos   cos 2  cos 3  0  (cos   cos 3 )  cos 2  0 U Example: 13 (d) 2n  E3 (a) (2n  1)  sin( A  B) cos A sin A 60 Example: 12 cos( A  B) sin A If  cos A  2 cos 2. cos   cos 2  0  cos 2 (2 cos   1)  0  Example: 14 2  4 or cos    1 2 2    2m    cos. 3 2 3 sin 6  sin 4  sin 2  0, then  equal to (a) n  or n   4 3 (b) n  or nπ  4 6 (c) n  or 2n   4 6 (d) None of these (sin 6  sin 2 )  sin 4  0  2 sin 4 cos 2  sin 4  0  sin 4 (2 cos 2  1)  0  sin 4  0 or or cos 2  cos 2 3 U Solution: (a)  2  2m    / 2    m   D YG  cos 2  0  cos 2 3 sin 4  sin 0 4  n  or 2  2n   n 4 or   n  ST    3 2 cos 2  1  0. 2.4 General Solution of Square of Trigonometrical Equations. (1) General solution of sin2 = sin2  : If sin 2   sin 2  or, 2 sin 2   2 sin 2  (Both the sides multiply by 2) or, 1  cos 2  1  cos 2 or, cos 2  cos 2 , 2  2n   2 ; n  I , θ  nπ  α; n  I (2) General solution of cos2  = cos2  : If cos 2   cos 2  or, 2 cos 2   2 cos 2  (multiply both the side by 2) or, 1  cos 2  1  cos 2 or, 2  2n  2 ; θ  nπ  α; n  I (3) General solution of tan2  = tan2: If tan 2   tan 2  or, tan 2  tan 2   1 1 Trigonometrical Equations and Inequations 51 tan 2   1 tan 2   1 Using componendo and dividendo rule,  tan 2   1 tan 2   1 1  tan 2  Example: 15  1  tan 2  1  tan 2  1  tan 2  1  tan 2   1  tan 2  1  tan 2  cos 2  cos 2 , θ  nπ  α; n  I or General value of  satisfying the equation tan 2   sec 2  1 is  (a) m  , n   Solution: (b) or (b) m  , n   3  tan 2   tan 2   sec 2  1  (c) m  , n   3 1  tan 2  1 1  tan 2  [IIT 1996]  6 60 1  tan 2  (d) None of these  tan 2   tan 4   1  tan 2   1  tan 2  E3 or tan 4   3 tan 2   0  tan 2  (tan 2   3)  0  tan 2   0 and tan 2   3  (b) n   6 2  6 (b) n    4 U 2 tan 2   1  tan 2   tan 2   1  tan 2   tan 2  If sin 2   sin 2   3 (d) n    3 If [MP PET 1989] (c) n    4  6 (d) 2n    4  4 1 , then the most general value of  is 4 1 1   4 2    n  Example: 19 . ST 4 (a) 2n   (1)n Solution: (c) (c) 2n   [MP PET 1988].  4    n  Example: 18. If 2 tan 2   sec 2  , then the general value of  is (a) n   Solution: (c) 3 3  3  4   cos 2   cos 2 sec    cos 2     3 6 4  2  2    n  Example: 17  6 D YG Solution: (b)  4 , then the general value of  is 3 If sec 2   (a) 2n   3    m  and   n   U Example: 16  ID tan 2   tan 2 0 and tan 2   tan 2  6 (b) n   (1)n 2 6 2  sin 2   sin 2  6. 1  cos 2  3, then the general value of  is 1  cos 2 [MP PET 1984, 90; UPSEAT 1973] (c) n    6 (d) 2n    6 52 Trigonometrical Equations and Inequations (a) 2n   (b) n   6  6 (c) 2n    (d) n   3  3 2 sin 2   1  cos 2  3  tan 2   3  ( 3 ) 2  tan 2   tan 2 3  3 1  cos 2 2 cos 2    n   3. 60 Solution: (d)  2.5 Solutions in the Case of Two Equations are given (Simultaneously Solving Equation). E3 We may divide the problem into two categories. (1) Two equations in one ‘unknown’ satisfied simultaneously. (2) Two equations in two ‘unknowns’ satisfied simultaneously. (1) Two equations is one ‘unknown’ : Two equations are given and we have to find the values of variables  which may satisfy with the given equations. θ  2nπ  α, n  I (i) cos   cos  and sin   sin  , so the common solution is θ  2nπ  α, n  I ID (ii) sin   sin  and tan   tan  , so the common solution is (iii) cos   cos  and tan   tan  , so the common solution is θ  2nπ  α, n  I The most general value of  satisfying the equation tan   1 and cos   U Example: 20 1990] Solution: (c) 7 4 (b) n   (1)n 7 4 D YG (a) n   1 2 is [MP PET 2003; UPSEAT 2002, 1982; Roorkee (c) 2n   7 4 (d) None of these 1     tan   1  tan  2   and cos    cos  2   4 4 2    7 . Hence, general value is 2n    2    2n   4 4   The most general value of  which will satisfy both the equations sin    U Example: 21 (a) n   (1)n sin   (b) n    6 [MNR 1980; MP PET 1989; DCE 1995] (c) 2n    6 (d) None of these 1       sin   sin    2 6  6   ST Solution: (d) 6 1 1 and tan   is 2 3  1 tan     3          tan    tan            6 6  6   Hence, general value of  is 2n   7. 6 (2) System of equations (Two equations in two unknowns) : Let f ( ,  )  0, g ( ,  )  0 be the system of two equations in two unknowns. Step (i) : Eliminate any one variable, say . Let    be one solution. Step (ii) : Then consider the system f ( ,  )  0, g ( ,  )  0 and use the method of two equations in one variable. Trigonometrical Equations and Inequations 53 Note :  It is preferable to solve the system of equations quadrant wise. 2  sin   tan    If   3, then the value of  and  are sin tan     (a)   n   (c)   n    3  2 ,   n  ,   n   (b)   n   6   3 ,   n  2    2 But,  sin  cos   sin  cos   sin 2  sin 2    2 E3  sin   tan     tan   sin   6 (d) None of these 3 2 Solution: (a)  60 Example: 22  tan  tan     3  tan 2   3    n   , so that   n   3  cot  3 6 tan  ID Trick: Check with the options for n  0, n  1. Example: 23 Solve the system of equations sec   2 sec , tan   3 tan  Solution: Usually students proceed this type of problems in the following way: U Squaring and subtracting, we get sec 2   tan 2   2 sec 2   3 tan 2  , i.e., 2 tan 2   2  3 tan 2   1 or tan 2   1 or .........(i) 4 sec 2  tan 2   2 3 D YG Also we have sec 2   tan 2     n  which gives 6  3 sec 2   2 tan 2  or tan 2   3, and so   m   Thus solution of this system is   m   Now see the fallacies:    3 and   n    4  3. , m, n  I........(ii)       and    (from the solution) give sec    2 sec    3 4 3  4 U     i.e., 2  2, but tan    3 tan    give 3    4 3   3. ST Thus solution given in (ii) consists many extraneous (absurd) solutions. The simple reason for this is quite obvious. (ii) consists of solutions of following four systems: sec   2 sec , tan   3 tan .........(iii) sec   2 sec , tan    3 tan .........(iv) sec    2 sec , tan   3 tan ..........(v) and sec    2 sec , tan    3 tan .........(vi) While we have to find the values which satisfy (iii). Therefore, we have to verify the solutions and should retain only the valid ones. Alternative Method : A better method for such type of equations is following: The given system is sec   2 sec ........(vii) tan   3 tan ........(viii) 54 Trigonometrical Equations and Inequations (vii)2 – (viii)2 gives tan 2   1    3 ,   2m    4 4 ,. 4.........(ix) 2 3. , then we have sec   2, tan    3 , so   4 3 2 3 ,   2m   3 4........(x) 3 5 (or can be taken as  ) 4 4 Then sec   2, tan   3 ,    Q 3 , so   4. 3 Thus   2n   3 4 Case 4:   4 , 3   2m   5 4 or   2n   2 , 3........(xi) 7  (or  ). 4 4  3 and so   2n    3 ,   2m    4...........(xii) U Then sec   2, tan    3 , so    ID   2m   ,. Thus general solution is   2n   Case 3 :   4   , the system reduces to sec   2, tan   3 , so  . 4 3  Case 2 :   , 60    2n   4 E3 Case 1 :    3 5 7 Hence, the required solutions are given as    ( ,  )   2n   , 2m    ; 3 4  2 3   , 2m    2n   , 3 4   D YG 2 3      , 2m    2n    ,  2n   , 2m   . 3 4 3 4    Note :  Do not write the solution as   2n    3 , 2n   2  ,................,   2m   ,.......... 3 4 2.6 General Solution of the form a cos + bsin = c. In a cos   b sin   c, put a  r cos  and b  r sin  where r  a 2  b 2 and | c |  a 2  b 2 U Then, r (cos  cos   sin  sin  )  c  cos(   )  c a  b2 2 ST      2n       2n      , where tan    cos  (say)..........(i) b , is the general solution a Alternatively, putting a  r sin  and b  r cos  where r  a 2  b 2  sin(   )  c a2  b 2  sin  (say)      n  (1)n     n  (1)n    , where tan   Note :  a , is the general solution. b ( a 2  b 2 )  a cos   b sin   ( a 2  b 2 )  c b  The general solution of a cos x  b sin x  c is x  2n   tan 1    cos 1  2 2 a  a b Example: 24  .   The number of integral values of k, for which the equation 7 cos x  5 sin x  2k  1 has a solution is [IIT Screeni (a) 4 (b) 8 (c) 10 (d) 12 Trigonometrical Equations and Inequations 55 Solution: (b)  7 2  5 2  (7 cos x  5 sin x )  7 2  5 2 So, for solution  74  (2k  1)  74 or 8.6  2k  1  8.6 or 9.6  2k  7.6 or 4.8  k  3.8. So, integral values of k are 4,  3,  2,  1, 0, 1, 2, 3 Example: 25 If (eight values) 3 cos   sin   2 , then general value of  is [MP PET 2002, 1991; UPSEAT 1999] 3 cos   sin   2   sin Example: 26 (b) (1)n 4  3 cos   cos  3  4   (c) n   3 2 If sin   cos   1 , then the general value of  is 1  (d) n   (1)n 3 sin   2  4   3 1 cos     4 1 2  [Karnataka CET 2002; DCE 2000; MNR 1987; IIT (c) 2n   4  (d) None of these 2 ID (b) n   (1)n sin   cos   1         sin    sin    n   (1)n . 3 4 4 3  1 1981] Solution: 4 3 1 1 cos   sin   2 2 2 sin   (a) 2n   60 Solution: (d)  E3 (a) n   (1)n Dividing by  2 12  12  2   Example: 27 U  1        sin     sin     n   (1)n    n   (1)n . 4 4 4 4 4 4 2  3 sin x  cos x  4 has The equation (b) Two solutions D YG (a) Only one solution (c) Infinitely many solutions Solution: (d) Given equation is [EAMCET 2001] (d) No solution 3 sin x  cos x  4 which is of the form a sin x  b cos x  c with a  3 , b  1, c  4 Here a  b  3  1  4  c 2 , Therefore the given equation has no solution. 2 Example: 28 2 The general solution of the equation ( 3  1) sin   ( 3  1) cos   2 is (a) 2n   4   (b) n   (1)n 12  4   12 (c) 2n    4   12 (d) n   (1)n  4   12 ( 3  1) sin   ( 3  1) cos   2  3  1   3  1 U Solution: (a)  [Roorkee 1992] Divided by ( 3  1) ST We get, 2 2 2 sin   2 ( 3  1) 2 2 sin  sin 15   cos  cos 15   1  2 2 in both sides, cos   2 2 2  sin . sin 2  12  cos . cos  12  cos  4         cos    2n        2n   .   cos 12  4 4 12 12 4  2.7 Some Particular Equations. (1) Equation of the form a 0 sin n x  a 1 sin n 1 x cos x  a 2 sin n 2 x cos 2 x ....  a n cos n x  0 : Here a0 , a1..., an are real numbers and the sum of the exponents in sin x and cos x in each term is equal to n, are said to be homogeneous with respect to sinx and cosx. For cos x  0, above equation can be written as, a0 tan n x  a1 tan n 1 x ...  an  0. 56 Trigonometrical Equations and Inequations The solution of equation 5 sin 2 x  7 sin x cos x  16 cos 2 x  4 is (a) x  n  tan 1 3 or x  n  tan 1 4 (c) x  n  or x  n   Solution: (a) (b) x  n     6 or x  n   (d) 4 4 None of these To solve this kind of equation; we use the fundamental formula trigonometrical identity, sin x  cos x  1 2  2 60 Example: 29 writing the equation in the form, 5 sin 2 x  7 sin x cos x  16 cos 2 x  4 (sin 2 x  cos 2 x )  sin2 x  7 sin x cos x  12 cos 2 x  0 E3 Dividing by cos 2 x on both sides we get, tan 2 x  7 tan x  12  0 Now it can be factorized as; (tan x  3)(tan x  4)  0  tan x  3, 4 i.e., tan x  tan(tan 1 3) or tan x  tan(tan 1 4 )  x  n  tan 1 3 or x  n  tan 1 4. ID (2) A trigonometric equation of the form R( sin kx, cos nx, tan mx, cot lx)  0 : Here R is a rational function of the indicated arguments and (k, l, m, n are natural numbers) can be reduced to a rational equation with respect to the arguments sin x , cos x , tan x , and cot x by means of the D YG U formulae for trigonometric functions of the sum of angles (in particular, the formulas for double and triple angles) and then reduce equation of the given form to a rational equation with x respect to the unknown, t  tan by means of the formulas, 2 x 1  tan 2 2 sin x  , cos x  2 x 1  tan 1  tan 2 2 2 tan Example: 30 1   If (cos x  sin x )  2 tan x    2  0 then x  cos x    (b) n   3 U (a) 2n    3 x Let t  tan , and using the formula. We get, 2 ST Solution: (a) x x x 2 tan 1  tan 2 2 , cot x  2 , tan x  2 x x x 1  tan 2 2 tan 2 2 2  1  t2 2 t    1  t2  1  t2     4t 1  t2  1  t2  1  t2  Its roots are; t1  1 3  20   and t2   1  (c) 2n    2  1  tan   1  tan 2   6 x x  2 tan 2  2  x x 1  tan 2  2 2  (d) None of these x  1  tan 2  4 tan 2    1  tan 2 x 1  tan 2  2 3 t 4  6 t 3  8 t 2  2t  3 0 (t 2  1) (1  t 2 ). 3 Thus the solution of the equation reduces to that of two elementary equations, tan x 1 x 1  , tan   2 2 3 3  x   n  2 6  x  2n    3 , is required solution. x 2   2  0 x  2  Trigonometrical Equations and Inequations 57 (3) Equation of the form R( sin x  cos x, sin x. cos x)  0 : where R is rational function of the arguments in brackets, Put sin x  cos x  t........(i) and use the following identity: (sin x  cos x )2  sin 2 x  cos 2 x  2 sin x cos x  1  2 sin x cos x  sin x cos x  t2 1 2 …….(ii) 60  t2 1 0. Taking (i) and (ii) into account, we can reduce given equation into; R  t, 2   Similarly, by the substitution (sin x  cos x )  t, we can reduce the equation of the form; If sin x  cos x  2 2 sin x cos x  0 then the general solution of x is  (a) x  2n   Solution: (c)  (b) x  n   (1)n 4 6   (c) Both (a) and (b) (d) None of these 4 ID Example: 31 E3  1  t2    0. R(sin x  cos x , sin x cos x )  0 to an equation; R  t,  2   Let (sin x  cos x )  t sin x. cos x  and using the equation The numbers t1  2 , t2   we get  t2 1 0 t  2 2    2   U 2t 2  t  2  0 t2 1 , 2 1 are roots of this quadratic equation. 2 D YG Thus the solution of the given equation reduces to the solution of two trigonometrical equation; sin x  cos x  2 or sin x  cos x   1 2 or or 1 sin x  2 2 sin x. cos  4 cos x  1 or  sin  4 1 x  2n   Example: 32  4 cos x  1 or sin x cos or x  n   (1)n If sin10 x  cos 10 x  (a) x  Solution: (b) Using 1 cos x   2   1   sin x    1 or sin x     4 4 2     ST  sin x  2 U  1  6  4  sin  x  4  4 1 2 cos x    (4 n  1)  2 1 2 or x   4    n   (1)n.   6    . 4 29 cos 4 2 x. then x  16 n 4 half-angle 5 (b) x  n   4 8 formulae we (c) x  can n   4 3 represent the (d) None of these given equation in the 5 29  1  cos 2 x   1  cos 2 x  cos 4 2 x      2 2 16     Put 5 5 1 t  1 t  29 4 t cos 2 x  t,       2   2  16  24 t 4  10 t 2  1  0 whose only real root is, t 2   cos 2 2 x  1  1  cos 4 x  1  cos 4 x  0  4 x  (2n  1)   x  n    ; n  I 2 2 4 8 1. 2 form, 58 Trigonometrical Equations and Inequations Note :  Some trigonometric equations can sometimes be simplified by lowering their degrees. If the exponent of the sines and cosines occuring into an equation are even, the lowering of the degree can be done by half angle formulas as in above example. 2.8 Method for Finding Principal Value. 1. 2 60 Suppose we have to find the principal value of  satisfying the equation sin    Since sin  is negative,  will be in 3rd or 4th quadrant. We can approach 3rd or 4th quadrant from two directions. If we take anticlockwise direction the numerical Y E3 value of the angle will be greater than . If we approach it in clockwise direction the angle will be numerically less than . For principal value, we have to take numerically smallest angle. So for principal value X ID (1) If the angle is in 1st or 2nd quadrant we must select anticlockwise direction and if the angle is in 3rd or 4th quadrant, we must select clockwise direction. /6 O X /6 A B Y U (2) Principal value is never numerically greater than . ST U D YG (3) Principal value always lies in the first circle (i.e., in first rotation). On the above criteria, 5     will be  or  has the least numerical value. Hence  is the. Among these two  6 6 6 6 1 principal value of  satisfying the equation sin   . 2 From the above discussion, the method for finding principal value can be summed up as follows : (i) First draw a trigonometrical circle and mark the quadrant, in which the angle may lie. (ii) Select anticlockwise direction for 1st and 2nd quadrants and select clockwise direction for rd 3 and 4th quadrants. (iii) Find the angle in the first rotation. (iv) Select the numerically least angle. The angle thus found will be principal value. (v) In case, two angles one with positive sign and the other with negative sign qualify for the numerically least angle, then it is the convention to select the angle with positive sign as principal value. Example: 33 If cos   3 sin   2, then   (only principal value) (a) Solution: (a) Example: 34 Solution: (a)  3 (b) 2 3 (c) 4 3 (d) 5 3  1 3 2      cos   sin    cos  cos  sin  sin  1  cos     1  cos 0      0    . 3 3 3 2 2 2 3 3  Principal value of tan   1 is   (a) (b) 4 4  tan  is negative. (c) 3 4 (d) 3 4 B Y X 3/4 – /4 X A Y Trigonometrical Equations and Inequations 59 Important Tips  60   will lie in 2nd or 4th quadrant. For 2nd quadrant we will select anticlockwise and for 4 th quadrant, we will select clockwise direction. 3  In the first circle two values and are obtained. 4 4   Among these two, is numerically least angle. Hence principal value is. 4 4 Any trigonometric equation can be solved without using any formula. Find all angles in 0, 2  which satisfy the For example: Consider the equation sin   E3 equation and then add 2n  to each.  5  5 1 , then   ,. Hence required solutions are   2n   , 2n  . 6 6 2 6 6 2.9 Important Points to be Taken in Case of While Solving Trigonometrical Equations. ID (1) Check the validity of the given equation, e.g., 2 sin   cos   4 can never be true for any  as the value (2 sin   cos  ) can never exceeds equation. 2 2  (1)2  5. So there is no solution to this U (2) Equation involving sec  or tan  can never have a solution of the form. (2n  1)  2 Similarly, equations involving cosec  or cot  can never have a solution of the form D YG   n . The corresponding functions are undefined at these values of . (3) If while solving an equation we have to square it, then the roots found after squaring must be checked whether they satisfy the original equation or not, e.g., Let x  3. Squaring, we get x 2  9  x  3 and  3 but x  3 does not satisfy the original equation x  3. e.g., sin x  cos x  1 Square both sides, we get 1  sin 2 x  1 or x U  2 x  n ST  Roots are ……,  sin 2 x  0 n , nI 2  3  2    2 3 , , ,0, , , ,...... 2 2 2 2 2 2 We find that 0 and  / 2 are roots but  and 3 / 2 do not satisfy the given equation as it leads to  1  1 Similarly 0 and   3 are roots but  and   are not roots as it will lead to  1  1. 2 2 As stated above, because of squaring we are solving the equations sin x  cos x  1 and sin x  cos x  1 both. The rejected roots are for sin x  cos x  1. (4) Do not cancel common factors involving the unknown angle on L.H.S. and R.H.S. because it may delete some solutions. e.g., In the equation sin  (2 cos   1)  sin  cos 2  if we cancel sin  on 60 Trigonometrical Equations and Inequations both sides we get cos 2   2 cos   1  0  (cos   1)2  0  cos   1    2n. But   n  satisfies the equation because it makes sin   0. So, the complete solution is   n , n  Z. also (5) Any value of x which makes both R.H.S. and L.H.S. equal will be a root but the value of x for which    will not be a solution as it is an indeterminate form. 60 Hence, cos x  0 for those equations which involve tan x and sec x whereas sin x  0 for those which involve cot x and cosec x. Also exponential function is always +ve and log a x is defined if x  0 , x  0 and a  0, a  1 (tan 2 x )  tan x and not  tan x. E3 f (x )  ve always and not  i.e. (6) Denominator terms of the equation if present should never become zero at any stage while solving for any value of  contained in the answer. (7) Sometimes the equation has some limitations also e.g., cot 2   cosec 2  1 can be true only ID if cot 2   0 and cosec 2  1 simultaneously as cosec 2  1. Hence the solution is   (2n  1) / 2. y z   y  z only and not x x x  0 , as it will make   . Similarly if ay  az , then it will also imply y  z only as a  0 being a (8) If xy  xz then x (y  z )  0  either x  0 or y  z or both. But U constant. Similarly x  y  x  z  y  z and x  y  x  z  y  z. Here we do not take x  0 as in the above D YG because x is an additive factor and not multiplicative factor. When cos   0 , then sin   1 or  1. We have to verify which value of sin  is to be chosen 1  which satisfies the equation. cos   0     n   . 2  (9) Student are advised to check whether all the roots obtained by them, satisfy the equation and lie in the domain of the variable of the given equation. U 2.10 Miscellaneous Examples. Example: 35 The equation 3 sin 2 x  10 cos x  6  0 is satisfied if ST 1 (a) x  n   cos 1   3 Solution: (b) 1 (b) x  2n   cos 1   3 1 1 (c) x  n   cos 1   (d) x  2n   cos 1   6 6 3 sin 2 x  10 cos x  6  0  3(1  cos 2 x )  10 cos x  6  0 on solving, (cos x  3)(3 cos x  1)  0. Either cos x  3 (which is not possible) or cos x  1 3  1 x  2n   cos 1  . 3 Example: 36 If the solutions for  of cos p  cos q  0, p  0, q  0 are in A.P., then the numerically smallest common difference of A.P. is (a)  pq [Kerala (Engg.) 2001] (b) 2 pq (c)  2( p  q) (d) 1 pq Trigonometrical Equations and Inequations 61 the solutions form an A.P.    Example: 37  p  2n  (  q ), n  I  Given, cos p    cos q   cos(  q  ) (2n  1) pq gives us an A.P. with common difference  tan 3 x  tan 2 x =1 is 1  tan 3 x tan 2 x 1993, 2002] (a)    4    (d) 2n   : n  1,2,3...  4    : n  1, 2,3...    Solution: (a) tan( 3 x  2 x )  tan x  1  x  n   Example: 38 If cos   cos 7  cos 3  cos 5  0, then   (b) n 2 (b) 4 U 1 n (sin 2 3 )  0  sin 8  0. Hence   8 2 sin   2 4 (2n  1) (c) 4 (2n  1) (d) None of these    tan x  2  tan(cot x )  tan  tan(cot x )  cot(tan x )   tan x U (2n  1) 2 ST 1 (2n  1)  sin x. cos x 2  sin x cos x (2n  1)   cos x sin x 2  sin 2 x  4. (2n  1)     1 The sum of all solutions of the equation cos x. cos   x . cos   x   , x  0,6  is 3  3  4 (a) 15 (b) 30  (c)  110  3 (d) None of these  3 cos x 1 1  1 4 cos 2 x  3  Here, cos x  cos 2 x  sin 2 x   or or cos 3 x  1 4 4 4 4  4  3 x  2n  x  Example: 41 (d) None of these 3 D YG  tan x  cot x  Solution: (b) n 8 (c) [ISM Dhanbad 1972] If tan(cot x )  cot(tan x ) , then sin 2 x equal to cot x  n   Example: 40. But this value does not satisfy the given equation. Combining  and 7 , 3 and 5θ, we get 2 cos 4 (cos 3  cos  )  0 (a) (2n  1) Solution: (b) 4 ID n 4  4 cos 4. cos 2. cos   0  4 Example: 39  4 [MP PET 1992; UPSEAT E3 (b)  (c) n   Solution: (c) and 2 2 (2n  1) 2 gives us an A.P. with common difference =. Certainly,.  pq pq pq pq The set of values of x for which the expression (a) 2 pq 60 Solution: (b) (2n  1) (2n  1) or , n  I. Both  pq pq 2n  , where n  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ; 3  The required sum  2 The equation sin x  sin y  sin z  3 for 0  x  2 ,0  y  2 ,0  z  2 , has (a) One solution (b) Two sets of solution 3 9  n = 30. n 0 [Orissa JEE 2003] 62 Trigonometrical Equations and Inequations (c) Four sets of solution (d) No solution 3 ; for x , y, z ,  0, 2 . 2 Solution: (a) Given, sin x  sin y  sin z  3 is satisfied only when x  y  z  Example: 42 The solution set of (5  4 cos  ) (2 cos   1)  0 in the interval 0, 2  is   (b)  ,   3  (5  4 cos  ) (2 cos   1)  0 cos   5 which is not possible 4 E3 Solution: (c)  2 5  (d)  ,   3 3   2 4   , (c)   3   3 60   2  (a)  ,  3 3  [EAMCET 2003] 2 4   2 cos   1  0 or cos    1    2 , 4. Solution set is  ,   0, 2  3 2 The equation 3 cos x  4 sin x  6 has Solution: (d) (b) Infinite solution 3 cos x  4 sin x  6  6 3 4 6  cos (x   )  cos x  sin x  5 5 5 5 3  (c) One solution [Orissa JEE 2002] (d) No solution ID (a) Finite solution  3  1  3     cos    5   U Example: 43 3 So that equation has no solution. Example: 44 The equation sin x  cos x  2 has D YG (a) One solution (b) Two solution (c) Infinite number of solution Solution: (d) [EAMCET 1986; MP PET 1998] (d) No solution No solution as sin x  1, cos x  1 and both of them do not attain their maximum value for the same angle. Trick: Maximum value of sin x  cos x  1 2  1 2  2 , Hence there is no x satisfying this equation. If 2 cos x  3 and x  ,   then the solution set for x is      (a)   ,    ,  6  6   (b)       ,  (c)   , 6   6   (d) None of these ST U Example: 45 Solution: (a) Here, cos x    , 6 6 0 3. The value scheme for this is shown below. 2 From the figure, cos x < 3/2 –1 3/2 1     x   x  or 6 6   x    ,  Example: 46        , . 6  6  3/2 0 The number of pairs (x, y) satisfying the equations sin x  sin y  sin( x  y ) and | x | | y |  1 is (a) 2 (b) 4 (c) 6 (d)  Trigonometrical Equations and Inequations 63 Solution: (c) 1 1 1 1 The first equation can be written as, 2 sin (x  y ) cos (x  y )  2 sin (x  y ) cos (x  y ) 2 2 2 2 1 2 1 2 1 2  Either sin (x  y )  0 or sin x  0 or sin y  0  x  y  0, or x  0 or y  0. As |x| + |y| =1, therefore when x  y  0, we have to reject x  y  1, or  1 1   1 1  ,  as the possible  ,  or  2 2   2 2  60 x  y  1 and solve it with x  y  1 or x  y  1 which gives solution. Again solving with x  0, we get (0,  1) and solving with y  0, we get (1,0) as the other If cos   1 and 0     360 , then the values of  are 2 (a) 120  and 300  Solution: (c) 1 2 Given, cos    cos 120    and 0    360 . We know that cos 60   (b) 2 2    sin  ) 1 (c) 2 D YG tan( cos  )  tan( U 1 2   sin   cos   1  cos      2 Example: 49 4 1 1 2 or 5 4 (b) 1 (d) 3 2 1 4 2. 2 2 The only value of x for which 2 sin x  2cos x  21(1 / (a) Solution: (a) and cos(180   60 )   cos 60     If tan( cos  )  cot( sin  ), then the value of cos     4  (a) Solution: (a) 1 2 (d) 60  and 240  1 1 1. Similarly cos (180 o  60 o )   cos 60    or cos 240  . 2 2 2 Therefore   120  and 240°. Example: 48 (c) 120  and 240  (b) 60  and 120  ID Example: 47 E3 solution. Thus we have six pairs of solution for x and y. 3 4 2) hold is (c)  2 (d) All values of x. Since A.M.  G.M. U 1 sin x  2 cos x )  2 sin x.2 cos x  2sin x  2cos x  2.2 (2 2 sin x  cos x 2 1  2sin x  2cos x  2 sin x  cos x 2 And, we know that sin x  cos x   2 ST  2 sin x  2 cos x  2 1(1 / 2) for x  5. 4 2.11 Periodic Functions. A function f(x) is called periodic function if there exists a least positive real number T such that f (x  T )  f (x ). T is called the period (or fundamental period) of function f (x ). Obviously, if T is the period of f (x ), then f (x )  f (x  T )  f (x  2T )  f (x  3 T ) ............ (i) If f1 (x ) and f2 ( x ) are two periodic functions of x having the same period T, then the function af1 (x )  bf 2 (x ) where a and b are any numbers, is also a periodic function having the same period T. 64 Trigonometrical Equations and Inequations (ii) If T is the period of the periodic function f (x ) , then the function f (ax  b), where a( 0) and b are any numbers is also a periodic function with period equal to T / a. (iii) If T1 and T 2 are the periods of periodic functions f1 (x ) and f2 ( x ) respectively, then the function af1 (x )  bf 2 (x ), where a and b are any numbers is also periodic and its period is T which 60 is the L.C.M. of T1 and T 2 i.e. T is the least positive number which is divisible by T1 and T 2. All trigonometric functions are periodic. The period of trigonometric function sin x , cos x , sec x and cosec x is 2 because sin( x  2 )  sin x , cos( x  2 )  cos x etc. The period of tan x and cot x is  because tan( x   )  tan x and cot( x   )  cot x . Here |a| is taken so as the value of the period is positive | a| real number. ID The period of tan ax and cot ax is 2 a E3 The period of the function which are of the type: sin ax, cos(ax  b); b cos ax is Some functions with their periods Function Period tan(ax  b), cot(ax  b) U sin(ax  b), cos(ax  b), sec(ax  b), cosec(ax  b) 2 / a  /a  /a | tan(ax  b) |, | cot(ax  b) |  /a Example: 50 D YG | sin(ax  b)|, | cos(ax  b)|, | sec(ax  b)|, | cosec(ax  b)| Period of sin 2 x is (a)  sin 2 x  (b) 2 1  cos 2 x 2  Period  U Solution: (a) Example: 51 (d) None of these ST [Kerala(Engg.) 2003] (c)  2 (d) 4 2 | a| 2  . | 2| The period of the function f ( )  sin (a) 3 Solution: (d)  2 2  . 2 (b)  Period of sin(ax  b)   Period of sin 2 x  Example: 52 (c) The period of the function y  sin 2 x is (a) 2 Solution: (a) [UPSEAT 2002; AIEEE 2002]  3  cos  2 is (b) 6     Period of sin   6 and period of cos    4  3 2 L.C.M. of 6 and 4  12 [EAMCET 2001] (c) 9 (d) 12 Trigonometrical Equations and Inequations The function f (x )  sin 2  2 cos (a) 6 Solution: (d) Period of sin Period of cos x 3  tan x 4 65 is periodic with period (b) x 2 x 3 3 (c)  2 4  /2  x 2   4  6 and period of tan 4  /4  /3  Period of f (x )  L.C.M. of (4, 6, 4)=12. (a) 1 (b) 4 (c) 8 Solution: (d) x x  1  sin   sin 2    sin  (2n   x ) n n  n  Example: 55 The period of sin 4 x  cos 4 x is (a) 12 [Pb. CET 2000] (d) 2 x  Period of the function sin  is 2n.  2n  4  n  2. n (c) 2 (b)  (sin2 x )2  (cos 2 x )2  1  2 sin 2 x cos 2 x  1  Therefore period is (d) (d) 3 2 1  1  cos 4 x  3 1 2 sin 2 2 x  1      cos 4 x 4 4   4 4 U Solution: (a)  2 4 E3 x If the period of the function f (x )  sin  is 4  , then n is equal to n ID Example: 54 [Rajasthan PET 2001] 60 Example: 53 x 2  . 4 2 D YG         Trick: f (x )  sin 4 x  cos 4 x  f   x   sin 4   x   cos 4   x   f   x   cos 4 x  sin 4 x  f (x ) 2 2 2 2         Hence the period is Example: 56 Period of sin   3 cos  is (a)  4 (b)  2 [MP PET 1990] (c)  (d) 2 1  3  . cos    2 sin   sin   3 cos   2  sin   2  3 2    U Solution: (d) . 2 Hence period  2. Period of | sin 2 x | is ST Example: 57 (a) Solution: (b)  4 [MP PET 1989] (b)  2 (c)  Period of sin 2 x   and period of | sin 2 x |   2. *** (d) 2