### Solve${\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x = 1$ &

Select correct option(s). x is odd multiple of ...

(A) $\frac {\pi}{6}$
(B) $\frac {\pi}{4}$
(C) $\frac {\pi}{3}$
(D) $\frac {\pi}{2}$

Solution

We have, ${\cos ^2}x + {\cos ^2}2x = {\sin ^2}3x$

$\Rightarrow ({\cos ^2}x - {\sin ^2}3x) + {\cos ^2}2x = 0$

$\Rightarrow \cos 4x.\cos 2x + {\cos ^2}2x = 0$

$\Rightarrow \cos 2x(\cos 4x + \cos 2x) = 0$

$\Rightarrow \cos 2x.2\cos 3x\cos x = 0$

$\Rightarrow \cos x.\cos 2x.\cos 3x = 0$

If cos x = 0, then x is odd multiple of $\frac {\pi}{2}$.

If cos 2x = 0, then x is odd multiple of $\frac {\pi}{4}$.

If cos 3x = 0, then x is odd multiple of $\frac {\pi}{6}$.