Question

Solve the following equation:
$3+5\sin 2x=\cos 4x.$

Answer 1

$3+5\left(\sin 2x\right)=1-{2\sin }^{2}2x$

Let $\sin 2x=z.$

$3+5z=1-{2z}^2$

${2z}^2+5z+2=0$

${2z}^2+4z+z+2=0$

$(2z+1)(z+2)=0$

So, $\sin 2x=-2$ is not possible.

$\therefore$ $\sin 2x=-\frac{1}{2},2x=n\pi +{(-1)}^n(-\frac{\pi }{6}),x=n\frac{\pi }{2}+{(-1)}^n(-\frac{\pi }{12}).$