Question

Solve the following equation:
$(\cos \frac{\pi }{4}-\sin \frac{\pi }{6})(\frac{1}{\cos x}+\tan x)=\sin \frac{\pi }{4}\cos x$

Answer 1

$⟹(\frac{1}{\sqrt{2}}-\frac{1}{2})\left(\frac{1+\sin x}{\cos x}\right)=\frac{1}{\sqrt{2}}\cos x$

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

$⟹2+2\sin x-\sqrt{2}-\sqrt{2}\sin x=2-\sqrt{2{\sin }^{2}x}=2-{2\sin }^{2}x$

Let $\sin x=z.$

${2z}^2+(2-\sqrt{2})z=0$

$2z=\sqrt{2}-2$

$z=\frac{1}{\sqrt{2}}-1=a$ $\left(Let\right)$

So, $\sin x=a,x=n\pi +{(-1)}^n{\sin }^{-1}a.$ and $\sin x=0,x=n\pi .$