Answer
Option C
Work Step by Step
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.
