Question

If $2\cos^{2}x\cos 2x = 0$then $x = (2n + 1)\frac{\pi}{2}\$

• A. $x = (2n + 1)\frac{\pi}{4}$
• B. $\therefore$
• C. $(2n \pm 1)\frac{\pi}{4}$
• D. $\frac{2\tan x + 1 - \tan^{2}x}{1 + \tan^{2}x} = \tan x$

Answer 1

Answer: (B)

$\therefore$

Answer 2

Eliminating $\sin\left( \frac{3x}{2} \right) = \frac{2\pi}{3/2} = \frac{4\pi}{3}$, we get $3\pi$ $\frac{4\pi}{3}$ (rejected)

$12\pi 12\pi$.