Answer
$\left\{\frac{\pi}{6} +\pi(n), \frac{\pi}{3}+\pi(n)\text{where n is any integer}\right\}$
Work Step by Step
RECALL:
$\sin{(2x)} = 2\sin{x}\cos{x}$
Thus, the given equation can be written as:
$2(2\sin{x}\cos{x})=\sqrt3
\\2(\sin{(2x))}=\sqrt3
\\\sin{(2x)}=\frac{\sqrt3}{2}$
RECALL:
$\sin{\theta}=a \longrightarrow \sin^{-1}{(a)}=\theta$
Thus,
$\sin{(2x)} = \frac{\sqrt3}{2} \longrightarrow 2x=\sin^{-1}{(\frac{\sqrt3}{2})}$
Use a scientific calculator's inverse sine function to obtain
$2x=\dfrac{\pi}{3} \text{ or } \dfrac{2\pi}{3}
\\x=\dfrac{\frac{\pi}{3}}{2}\text{ or } \dfrac{\frac{2\pi}{3}}{2}
\\x=\dfrac{\pi}{6} \text{ or } \dfrac{\pi}{3}$
The period of the function $\sin{(2x)}$ is $\pi$.
This means that the function's value repeats every interval of $\pi$.
Thus, if $\frac{\pi}{6}$ is a solution of the given equation, then $\frac{\pi}{6} + \pi(n)$, where $n$ is any integer, are all solutions of the equation.
Similarly, if $\frac{\pi}{3}$ is a solution, then $\frac{\pi}{3}+\pi(n)$ where $n$ is any integer are all solutions of the equation.
Thus, the solutions to the given equation are:
$\left\{\frac{\pi}{6} +\pi(n), \frac{\pi}{3}+\pi(n)\text{where n is any integer}\right\}$
