Answer
$x=0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{ 3}$
Work Step by Step
We have $\tan x +\pi =\dfrac{\tan x+\tan \pi}{1-\tan x \tan \pi} =\tan x$
$2 \sin x +\pi =2 (\sin x \cos \pi +\cos x \sin \pi) =2(-\sin x)$
After simplifying, we get $\tan x -2 \sin x =0$
or, $ \dfrac{\sin x}{\cos x} =2 \sin x$
or, $\sin x(2 \cos x -1)=0$
When $\sin x=0 \implies x=0, \pi$
and when $2 \cos x -1=0 \implies x= \dfrac{\pi}{3}, \dfrac{5\pi}{ 3}$
Thus, we have: $x=0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{ 3}$
