Solutions of Trigonometric Equations PDF

version: 1.1 CHAPTER 14 Solutions of Trignometric Equation 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 14. Solutio...

version: 1.1 CHAPTER 14 Solutions of Trignometric Equation 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab Example 2: Solve the equation: 1 + cos x = 0 14.1 Introduction Solution: 1 + cos x = 0 ⇒ cos x = -1 The Equations, containing at least one trigonometric function, are called Trigonometric Since cos x is -ve, there is only one solution x = p in [0, 2p] Equations, e.g., each of the following is a trigonometric equation: Since 2p is the period of cos x = ,= - x=+ sin 2 x sec x 1 ∴ General value of x is p + 2np, n∈Z 2 3 sin x Sec x tan and Hence solution set = {p + 2np}, n∈Z 5 4 Trigonometric equations have an ininite number of solutions due to the periodicity of the Example 3: Solve the equation: 4 cos2x - 3 = 0 trigonometric functions. For example If sin q = q then q = 0, ± , ± 2 ,... Solution: 4 cos2 x - 3 = 0 = as q ∈n , where n Z. ⇒ cos 2 x =⇒ ± cos x = which can be written 3 3 4 2 In solving trigonometric equations, irst ind the solution over the interval whose If cos x = 3 length is equal to its period and then ind the general solution as explained in the following i. 2 Since cos x is +ve in I and IV Quadrants with the reference angle examples: Example 1: Solve the equation sin x = 1 2 x= where x ∈ [ 0, 2 ] 6 sin x = =∴= -and x = 2 1 11 Solution: x 6 6 6 As 2p is the period of cos x. 2 sin x is positive in I and II Quadrants with the reference angle x =. ∴ + 2n + 2n , n∈Z a 11 where x ∈ [ 0, 2 ] 6 General value of x are and = ∴ x and =x= - 5 6 6 , 6 6 6 if cos x = - 3 ii. ∴ General values of x are + 2n , n ∈ Z 2 Since cos x is -ve in II and III Quadrants with reference angle x = 5 + 2n and 6 6   5  where x ∈ [ 0, 2 ] 6 Hence solution set = + 2n  ∪  + 2n  ,n ∈ Z ∴ x= - = and x = x + = 6  6  5 7 As 2p is the period of cos x. 6 6 6 6 version: 1.1 version: 1.1 2 3 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab ∴ General values of x are + 2n , n ∈ Z sin x cos x = 5 7 3 + 2n and Example 2: Find the solution set of:. 6 6 4   11  5  7  sin x cos x = 3 set = + 2n  ∪  + 2n  ∪  + 2n  ∪  + 2n  Solution:. 6   6  6  6  Hence solution 4 ⇒ ( 2sin x cos x ) = 1 3 2 4 ⇒ sin 2 x = 14.2 Solution of General Trigonometric Equations 3 2 When a trigonometric equation contains more than one trigonometric functions, trigonometric identities and algebraic formulae are used to transform such trigonometric a sin 2x is +ve in I and II Quadrants with the reference angle 2 x = 3 equation to an equivalent equation that contains only one trigonometric function. ∴ 2x = and 2 x = - = are two solutions in [ 0,2 ] The method is illustrated in the following solved examples: 2 3 3 3 As 2p is the period of sin 2x. Example 1: Solve: sin x + cos x = 0. ∴ + 2n and + 2n , , Solution: sin x + cos x = 0 2 General values of 2x are nU Z ⇒ + = ( Dividing by cos x ≠ 0 ) 3 3 sin x cos x 0 ⇒ General values of x are +n +n cos x cos x ⇒ tan x + 1 = 0 ⇒ tan - x = 1 and , nU Z tan x is -ve in II and IV Quadrants with the reference angle 6 3     Hence solution set = =  + n  ∪  + n  a 6  3  , nU Z x= where x ∈ [ 0, ] 4 Note: In solving the equations of the form sin kx = c, we irst ind the solution pf sin u = c ∴ x= - = (where kx = w) and then required solution is obtained by dividing each term of this 3 , As p is the period of tan x, 4 4 solution set by k. ∴ +n , 3 Example 3: Solve the equation: sin 2x = cos 2x General value of x is nU Z 4 3  ∴ Solution set =  + n  Solution: sin2x = cos2x 4  ⇒ ,n U Z. 2sinx cos x = cosx ⇒ 2sinx cos x - cosx = 0 ⇒ cosx(2sinx - 1) = 0 version: 1.1 version: 1.1 4 5 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab ∴ cosx = 0 or 2sinx - 1 = 0 As 2p is the period of cos x p 3p ∴ General values of x are + 2np and + 2np , nUZ i. If cosx = 0 ⇒ x= x= [0,2 p] 2 2 3 and where x U 2 2 ⇒ x = 0 and x = 2p where x U [0, 2p] ii. If cos x = 1 As 2p is the period of cos x. As 2p is the period of cos x , p 3p ∴ General values of x are + 2np and + 2np, nUZ, ∴ General values of x are 0 + 2np and 2p + 2np, nUZ. If 2 sin x - 1 = 0 2 2 p   3p + 2n p  ∪ {2np } ∪ {2p + 2np } , n ∈ z  ii. ∴ Solution Set = + 2n p  ∪  ⇒ 2   2  1 sin x = {2(n + 1)p } ⊂ {2np } , n ∈ z p 2 Since sin x is +ve in I and II Quadrants with the reference angle x = p p 5p where x U [0, 2p] 6 ∴ x= p   3p + 2n p  ∪ {2n p } , n ∈ z and x = p - =  Hence the solution set =  + 2np  ∪  2   2  6 6 6 As 2p is the period of sin x. Sometimes it is necessary to square both sides of a trigonometric equation. In such p p ∴ General values of x are and + 2np and 5 + 2np, nUZ, a case, extaneous roots can occur which are to be discarded. So each value of x must be 6 6 checked by substituting it in the given equation. For example, x = 2 is an equation having a root 2. On squaring we get x2 - 4 which gives p   3p  p   p  Hence solution set =  + 2np  ∪  + 2n p  ∪  + 2n p  ∪ 5 + 2n p  , two roots 2 and -2. But the root -2 does not satisfy the equation x = 2. Therefore, -2 is an 2   2  6   6  n∈ z extaneous root. Example 4: Solve the equation: sin2 x + cos x = 1. Example 5: Solve the equation: csc= x 3 + cot x. Solution: sin2 x + cos x = 1 Solution: csc= x 3 + cot x.......(i) ⇒ 1 - cos2 x + cos x = 1 ⇒ - cos x (cos x - 1) = 0 ⇒ = 3+ 1 cos x ⇒ or cos x - 1 = 0 sin x sin x ⇒ = 3 sin x + cos x cos x = 0 1 ⇒ 1 - cos x =3 sin x i. If cos x = 0 p 3p ⇒x= and x = where x U [0, 2p] ⇒ (1 - cos x) 2 = , 2 2 ( 3 sin x) 2 version: 1.1 version: 1.1 6 7 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab 14. Solutions of Trigonometric Equations eLearn.Punjab ⇒ 1 - 2cos x + cos 2 x = 3sin 2 x Exercise 14 ⇒ 1 - 2cos x + cos 2 x = 3(1 - cos 2 x) 1. Find the solutions of the following equations which lie in [0, 2p] ⇒ 4cos 2 x - 2cos x - 2 = sin x = - sec x = -2 cot q = 0 3 1 i) ii) cosecq = 2 iii) iv) ⇒ 2cos 2 x - cos x - 1 = 2 3 0 ⇒ (2cos x + 1)(cos x - 1) = 2. Solve the following trigonometric equations: 0 tan 2 q = ii) cos ec 2q = sec 2 q = cot 2 q = 1 4 4 1 ⇒ cos - x = or = 1 i) iii) iv) cos x 1 3 3 3 3 Find the values of q satisfying the following equations: 2 If cos x = - 3tan 2 q + 2 3 tan q + 1 = 1 3. tan q - secq - 1 = i. 0 2 2 4. 2sin q + cos q - 1 = 0 p Since cos x is -v e in II and III Quadrants with the reference angle x = 2 5. 2sin q - sin q = 0 2 6. p 2p p 4p 0 3cos 2 q - 2 3 sin q cosq - 3sin 2 q = 3 ⇒ x =p - = and x =p + = , where x U [0, 2p] 7. 0 [Hint: Divide by sin2q] 3 3 3 3 Find the solution sets of the following equations: 4 sin2q - 8cosq + 1 = 0 4p 8. Now x = does not satisfy the given equation (i). 9. 3 tan x - sec x - 1 =0 [Hint: sin3x = 3sinx - 4sin3x] 3 10. cos 2x = sin 3x 4p 2p ∴ x = is not admissible and so x = 11. sec 3q= secq is the only solution. 12. tan 2q + cotq = 0 3 3 Since 2p is the period of cos x 13. sin 2x + sinx = 0 14. sin 4x - sin 2x = cos 3x 2p ∴ General value of x is + 2n p , nUZ 15. sin x + cos 3x = cos 5x 16. sin 3 x + sin 2x + sin x = 0 sin 7x - sin x = sin 3 x 3 and x = 2p where x U [0, 2p] ii. If cos x = 1 ⇒ x= 0 17. 18. sin x + sin 3x + sin 5x = 0 sin q + sin 3q + sin 5q + sin 7q = 0 Now both csc x and cot x are not defined for x = 0 and x = 2 ∴ x= 0 19. cos q + cos 3q + cos 5 q + cos 7q = 0 and x = 2 are not admissible. 20.  2p  Hence solution set =  + 2n p  , nUZ  3  version: 1.1 version: 1.1 8 9