Question

Solve the following equation:
$\sin x-2\cos 2x=1$

Answer 1

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

${4\sin }^{2}x+\sin x-3=0$

$(4\sin x-3)(\sin x+1)=0$

So, $\sin x=\frac{3}{4},x=n\pi +{(-1)}^n{sin}^{-1}\left(\frac{3}{4}\right).$

And, $\sin x=-1,x=2n\pi +\frac{\pi }{2}.$