Question

The general solution of the following equation : $2(\sin x-\cos 2 x)-\sin 2 x(1+2 \sin x)+2 \cos x=0$ is/are -

• A. $x =2 n \pi ; n \in I$
• B. $n \pi+(-1)^{ n }\left(-\frac{\pi}{2}\right) ; n \in I$
• C. $x = n \pi+(-1)^{ n } \frac{\pi}{6} ; n \in I$
• D. $x = n \pi+(-1)^{ n } \frac{\pi}{4} ; n \in I$

Answer 1

Answer: (A), (B), (C)

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