Question

Solve the equation
$\left(viii\right)\sin x\tan x-1=\tan x-\sin x$

Answer 1

$\sin x\tan x-1=\tan x-\sin x$

$\Rightarrow \sin x\tan x+\sin x-1-\tan x=0$

$\Rightarrow \sin x(\tan x+1)-1(1+\tan x)=0$

$\Rightarrow (\sin x-1)(\tan x+1)=0$

$\Rightarrow (\sin x-1)=0\text{or}(\tan x+1)=0$

$\Rightarrow \sin x=1\text{or}\tan x=-1$

$\Rightarrow \sin x=\sin \frac{\pi }{2}\text{or}\tan x=\tan \frac{3\pi }{4}$

We know the general solution of $\sin x=\sin \alpha$ is $x=n\pi +{(-1)}^n\alpha ,n\in Z$

We know the general solution of $\tan x=\tan \alpha$ is $x=m\pi +\alpha ,m\in Z$

Hence, the general solution of the given eqution is
$x=n\pi +{(-1)}^n\frac{\pi }{2}\text{and}x=m\pi +\frac{3\pi }{4},n,m\in Z$