Answer
The solutions of the equation are $n\pi $ or $x=\pi +2n\pi $, where n is any integer.
Work Step by Step
The x in $\sin x=0$. Because $\sin 0=0$, and the solution of $\sin x=0$ in [0, $2\pi $ ) is:
$x=0$
$\begin{align}
& x=\pi +0 \\
& =\pi
\end{align}$
The period of the tangent function is $2\pi $, and the solutions are given by:
$x=0+n\pi =n\pi $ or $x=\pi +2n\pi $,
where n is any integer.
