Question

Find the general solutions of the following equations;
$2(\sin x-\cos 2x)-\sin 2x(1+2\sin x)+2\cos x=0$.

Answer 1

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

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

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

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

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

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

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

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

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

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

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

$\sin x=\frac{-1+\sqrt{1-4\times 2(-1)}}{2\times 2}$

$\sin x=-1$

$\sin x=\frac{1}{2},\sin x=-1,\cos x=1$