Question

Find the general solution for the equation $cos4x = cos2x$.

• A. $\frac{n\pi}{2}$, $n \in Z$
• B. $n\pi$, $n \in Z$
• C. $\frac{n\pi}{3}$, $n \in Z$
• D. Both $(b)$ and $(c)$

Answer 1

Answer: (D)

Both $(b)$ and $(c)$

Answer 2

We have, $cos4x = cos2x$ $\therefore$ The general solution is $4x = 2n\pi ? 2x$ $\left[\because cos\theta=cos\alpha \Rightarrow \theta=2n\pi \pm\alpha, n \in Z\right]$ $\Rightarrow 4x = 2n\pi + 2x$ or $4x = 2n\pi - 2x$ $\Rightarrow 4x - 2x = 2n\pi$ or $4x + 2x = 2n\pi$ $\Rightarrow 2x=2n\pi$ or $6x=2n\pi$ $\Rightarrow x=n\pi$ or $x=\frac{1}{3} n\pi$, where $n \in Z$ Hence, the required general solution is $x=\frac{n\pi}{3}$, $n \in Z$.