Answer
The solutions of the equation are $x=\frac{4\pi }{3}+2n\pi $ or $x=\frac{5\pi }{3}+2n\pi $, where n is any integer.
Work Step by Step
$\begin{align}
& 2sinx+\sqrt{3}=0 \\
& 2sinx=0-\sqrt{3} \\
& sinx=-\frac{\sqrt{3}}{2}
\end{align}$
$sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}$, and the solutions of $sinx=-\frac{\sqrt{3}}{2}$ in [0,2 $\pi $ ) are:
$\begin{align}
& x=\pi +\frac{\pi }{3} \\
& =\frac{3\pi }{3}-\frac{\pi }{3} \\
& =\frac{4\pi }{3}
\end{align}$
$\begin{align}
& x=2\pi -\frac{\pi }{3} \\
& =\frac{6\pi }{3}-\frac{\pi }{3} \\
& =\frac{5\pi }{3}
\end{align}$
The period of the sine function is $2\pi $, and the solutions are given by:
$x=\frac{4\pi }{3}+2n\pi $ or $x=\frac{5\pi }{3}+2n\pi $,
where n is any integer.
