## Algebra 2 (1st Edition)

Here, we have $\sin (x- 2\pi) +tan (x-2 \pi)=0$ The periodicity for $\sin x$ is $2 \pi$ and the periodicity for $\tan x$ is $\pi$. $\sin x +\tan x=0 \implies \sin x+\dfrac{\sin x}{\cos x}=0$ or, $\sin x(1+\dfrac{1}{\cos x})=0$ Thus, $\sin x=0$ or, $(1+\dfrac{1}{\cos x})=0$ We can see that the interval for $\sin x$ is $0$ when $x= 2\pi$ Hence, Option C is correct.