## Precalculus (6th Edition) Blitzer

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.
\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.