## Algebra 2 (1st Edition)

$x=0, \pi, \dfrac{\pi}{3}, \dfrac{ 5\pi}{3}$
Here, we have $\tan (x+\pi) + 2\sin (x+\pi)=0$ Use formula: $\tan (x+y) = \dfrac{\tan x+\tan y}{1-\tan x \tan y}$ $\dfrac{\tan x+\tan \pi}{1-\tan x \tan \pi} + 2\sin x \cos \pi+2 \cos x \sin \pi=0$ $\tan x- 2 \sin x=0 \implies \dfrac{\sin x}{\cos x} -2 \sin x=0$ or, $\sin x(\dfrac{1}{\cos x}-2)=0$ or, $(\sec x-2)\sin x=0$ a) When $\sin x=0$ Then in the interval $0 \leq x \leq 2 \pi$, we have $x= 0, \pi$ b) When $(\sec x-2)=0 \implies \sec x=2$ Thus, in the interval $0 \leq x \leq 2 \pi$, we have $x=\dfrac{\pi}{3}, \dfrac{ 5\pi}{3}$ Hence, $x=0, \pi, \dfrac{\pi}{3}, \dfrac{ 5\pi}{3}$