Question

Solve the following equation:
${\sin }^{2}x(1+\tan x)=3\sin x(\cos x-\sin x)3$

Answer 1

${\sin }^{2}x(1+\tan x)=3\sin x(\cos x-\sin x)3$

$\sin x=0,x=n\pi .$

and,

$\sin x(\cos x+\sin x)=9\cos x(\cos x-\sin x)$

$\sin x\cos x+{\sin }^{2}x=9{\cos }^{2}x-9\sin x\cos x$

${\sin }^{2}x+10\sin x\cos x-9{\cos }^{2}x=0$

${\tan }^{2}x+10\tan x-9=0$

$\tan x=\frac{-10\pm \sqrt{{10}^2+36}}{2}=\frac{-10\pm \sqrt{136}}{2}=\frac{-10\pm \left(\sqrt{34}\right)\times 2}{2}$

$\tan x=-5\pm \sqrt{36}$

Let $-5+\sqrt{36}=a,-5-\sqrt{36}=b$

So, $x=n\pi +{\tan }^{-1}a$ and $x=n\pi +{\tan }^{-1}b$