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. $x=2nÏ€;n∈I$
• B. $nÏ€+(−1{)}^n\left(−\frac{Ï€}{2}\right);n∈I$
• C. $x=nÏ€+(−1{)}^n\frac{Ï€}{6};n∈I$
• D. $x=nÏ€+(−1{)}^n\frac{Ï€}{4};n∈I$

Answer 1

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

$2(\sin x−\cos 2x)−\sin 2x(1+2\sin x)+2\cos x−0$
$⇒2\sin x−\sin 2x−2\cos 2x−2\sin x\sin 2x+2\cos x=0$
$⇒2\sin x−\sin 2x−2\cos 2x−(\cos x−\cos 3x)+2\cos x=0$
$⇒2\sin x(1−\cos x)+4{\cos }^{3}x−3\cos x+\cos x−2\left(2{\cos }^{2}x−1\right)=0$
$⇒2\sin x(1−\cos x)+4{\cos }^{3}x−4{\cos }^{2}x−2\cos x+2=0$
$⇒2\sin x(1−\cos x)−4{\cos }^{2}x(1−\cos x)$
$⇒+2(1−\cos x)=0$
$⇒(1−\cos x)\left\{2\sin x−4\left(1−{\sin }^{2}x\right)+2\right\}=0$
$⇒x=2nπ(2\sin x−1)(\sin x+1)=0$
$\sin x=\frac{1}{2}$ or $\sin x=−1$
$x=2nπ,x=nπ+(−1{)}^n\frac{π}{x}$ or $x=2nπ−\frac{π}{0}$