Question

Solve the following equation:
$\cos 2x+3\sin x=2$

Answer 1

$\cos 2x+3\sin x=2$

Put $\cos 2x=1-{2\sin }^{2}x$ and we get

${2\sin }^{2}x-3\sin x+1=0$

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

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

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

Either, $\sin x=1,$

$x=2n\pi +\frac{\pi }{6}.$

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