Question

If $3 \cos x \neq 2 \sin x$, then the general solution of $\sin ^{2} x-\cos 2 x=2-\sin 2 x$ is

• A. $n \pi+(-1)^{n} \frac{\pi}{2}, n \in Z$
• B. $\frac{n \pi}{2}, n \in Z$
• C. $(4 n \pm 1) \frac{\pi}{2}, n \in Z$
• D. $(2 n-1) \pi, n \in Z$

Answer 1

Answer: (C)

$\sin ^{2} x-\cos 2 x=2-\sin 2 x$
$\Rightarrow 1-\cos ^{2} x-\left(2 \cos ^{2} x-1\right)$
$=2-2 \sin x \cos x$
$\Rightarrow -3 \cos ^{2} x+2 \sin x \cos x=0$
$\Rightarrow \cos x(2 \sin x-3 \cos x)=0$
$\Rightarrow \cos x=0,(\because 2 \sin x-3 \cos x \neq 0)$
$\Rightarrow x=2 n \pi \pm \frac{\pi}{2}$
$\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}$