Question

Find the general solution of the following equations:
$2(\sin x-\cos 2x)-\sin 2x(1+2\sin x)+2\cos x=0$

Answer 1

$2(\sin x-\cos 2x)-\sin 2x(1+2\sin x)+2\cos x=0$

Simplyfing & expanding

$\sin x-\cos 2x-\sin x\cos x-{\sin }^{2}x\cos x+\cos x=0$

$\sin x-(1-2{\sin }^{2}x)-\sin x\cos x-{\sin }^{2}x\cos x+\cos x=0$

$[\because {\cos }^{2}x=1-{\sin }^{2}x,\sin 2x=2\sin x\cos x]$

$\sin x-1+2{\sin }^{2}x-\sin x\cos x-{\sin }^{2}x\cos x+\cos x=0$

$\sin x-\sin x\cos x+2{\sin }^{2}x-{\sin }^{2}2x\cos x+\cos x-1=0$

$\sin x(1-\cos x)+2{\sin }^{2}x(1-\cos x)-(1-\cos x)=0$

$(1-\cos x)(2{\sin }^{2}x+\sin x-1)=0$

$\cos x=1$ or $2{\sin }^{2}x+\sin x-1=0$

$(\sin x+1)(\sin x-\frac{1}{2})=0$

$cosx=1x=2n\pi$

$\sin x=-1x=n\pi +(-1{)}^n\frac{3\pi }{2}n\in I$

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