Question

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

• A. $x=2n\pi ;n\in I$
• B. $n\pi +{(-1)}^n(-\frac{\pi }{2});n\in I$
• 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)

$x=2n\pi ;n\in I$
$x=n\pi +{(-1)}^n\frac{\pi }{6};n\in I$
$n\pi +{(-1)}^n(-\frac{\pi }{2});n\in I$

$2(\sin x-\cos 2x)-\sin 2x(1+2\sin x)+2\cos x=0$
$\Rightarrow 2\sin x-\sin 2x-2\cos 2x-2\sin x\sin 2x+2\cos x=0$
$\Rightarrow 2\sin x-\sin 2x-2\cos 2x-(\cos x-\cos 3x)+2\cos x=0$
$\Rightarrow 2\sin x(1-\cos x)+4{\cos }^{3}x-3\cos x+\cos x-2(2{\cos }^{2}x-1)=0$
$\Rightarrow 2\sin x(1-\cos x)+4{\cos }^{3}x-4{\cos }^{2}x-2\cos x+2=0$
$\Rightarrow 2\sin x(1-\cos x)-4{\cos }^{2}x(1-\cos x)+2(1-\cos x)=0$
$\Rightarrow (1-\cos x)\{2\sin x-4(1-{\sin }^{2}x)+2\}=0$
$\Rightarrow \cos x=1$ or $\sin x-2(1-{\sin }^{2}x)+1=0$
$\Rightarrow x=2n\pi$ or $(2\sin x-1)(\sin x+1)=0$ $\Rightarrow \sin x=\frac{1}{2}$ or $\sin x=-1$
Hence solution set is $x=2n\pi ,x=n\pi +{(-1)}^n\frac{\pi }{6}$ or $x=n\pi -(-1{)}^n\frac{\pi }{2}$