Cos3x=4cos³x-3cosx
Understand the Problem
The question is asking us to solve the equation Cos3x=4cos³x-3cosx, which involves trigonometric identities and possibly finding the values of x that satisfy this equation.
Answer
The solutions for $x$ are: $$ x = \frac{\pi}{2} + k\pi, \quad x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{11\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad x = \frac{7\pi}{6} + 2k\pi $$ where $k$ is any integer.
Answer for screen readers
The general solutions for $x$ are:
$$ x = \frac{\pi}{2} + k\pi, \quad x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{11\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad x = \frac{7\pi}{6} + 2k\pi $$
where $k$ is any integer.
Steps to Solve
- Identify the equation We start with the equation
$$ \cos(3x) = 4\cos^3(x) - 3\cos(x) $$
This is a trigonometric identity. The right-hand side represents the cosine of a triple angle.
- Use the triple angle identity Recognize that
$$ \cos(3x) = 4\cos^3(x) - 3\cos(x) $$
is an identity. Since both sides are equal, we can proceed to find solutions for $x$.
- Set the equation to zero Rearranging the equation, we get:
$$ 4\cos^3(x) - 3\cos(x) - \cos(3x) = 0 $$
- Substituting $y$ for $\cos(x)$ Let ( y = \cos(x) ). Then we replace $\cos(x)$ in our equation:
$$ 4y^3 - 3y - \cos(3x) = 0 $$
- Finding roots Now we need to find $y$ values that satisfy
$$ 4y^3 - 3y = 0 $$
Factoring out $y$, we get:
$$ y(4y^2 - 3) = 0 $$
- Solve for $y$ Setting each factor to zero gives:
- ( y = 0 )
- ( 4y^2 - 3 = 0 )
For ( 4y^2 - 3 = 0 ):
$$ 4y^2 = 3 \implies y^2 = \frac{3}{4} \implies y = \pm\frac{\sqrt{3}}{2} $$
- Finding $x$ from $y$ values Now we translate back to $x$ using ( y = \cos(x) ):
- For ( y = 0 ), ( \cos(x) = 0 ) gives ( x = \frac{\pi}{2} + k\pi ) for integers ( k )
- For ( y = \frac{\sqrt{3}}{2} ), ( \cos(x) = \frac{\sqrt{3}}{2} ) gives ( x = \frac{\pi}{6} + 2k\pi ) or ( x = \frac{11\pi}{6} + 2k\pi )
- For ( y = -\frac{\sqrt{3}}{2} ), ( \cos(x) = -\frac{\sqrt{3}}{2} ) gives ( x = \frac{5\pi}{6} + 2k\pi ) or ( x = \frac{7\pi}{6} + 2k\pi )
The general solutions for $x$ are:
$$ x = \frac{\pi}{2} + k\pi, \quad x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{11\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad x = \frac{7\pi}{6} + 2k\pi $$
where $k$ is any integer.
More Information
This equation is a demonstration of the trigonometric identities, specifically involving the cosine function. The identity used here helps simplify complex trigonometric expressions and can be derived using angle addition formulas.
Tips
- Confusing the steps in solving for $y$ and returning to $x$ without keeping track of the correct quadrants for cosine.
- Failing to recognize the multiple angles can yield multiple values for $x$ due to the periodic nature of the cosine function.
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