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)

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