Answer
The solutions of the equation are $x=\frac{5\pi }{6}+2n\pi $ or $x=\frac{7\pi }{6}+2n\pi $, where n is any integer.
Work Step by Step
$\begin{align}
& 2\cos x+\sqrt{3}=0 \\
& 2\cos x=0-\sqrt{3} \\
& \cos x=-\frac{\sqrt{3}}{2}
\end{align}$
$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$, and the solutions of $\cos x=-\frac{\sqrt{3}}{2}$ in [0,2 $\pi $ ) are:
$\begin{align}
& x=\pi -\frac{\pi }{6} \\
& =\frac{6\pi }{6}-\frac{\pi }{6} \\
& =\frac{5\pi }{6}
\end{align}$
$\begin{align}
& x=\pi +\frac{\pi }{6} \\
& =\frac{6\pi }{6}+\frac{\pi }{6} \\
& =\frac{7\pi }{6}
\end{align}$
The period of the sine function is $2\pi $, and the solutions are given by:
$x=\frac{5\pi }{6}+2n\pi $ or $x=\frac{7\pi }{6}+2n\pi $,
where n is any integer.
