General solution to trigonometric equation $2\sin(3x) = \sqrt{3}$ After finding the trigonòmetric eqaution can someone help me find the values of x as I'm not really sure what to do 
Find the values of $x$ for which $2\sin(3x)= \sqrt{3}$.
 A: Ok, so you want to solve $\sin 3x = \frac{\sqrt{3}}{2}$.
Letting $\alpha = 3x$ you can rewrite $\sin \alpha = \frac{\sqrt{3}}{2}$, which is an elementary equation you can solve using tables: in fact, $\sin \alpha = \frac{\sqrt{3}}{2}$ iff $\alpha = 60^\circ \lor \alpha = 120^\circ$. Since the values of the sine of an angle repeat themselves every $360^\circ$ (i.e., $\sin \alpha$ is a $360^\circ$-periodic function), each angle in the form:
$$\tag{A} \alpha = 60^\circ + k 360^\circ \lor \alpha = 120^\circ + k 360^\circ$$
 (with $k \in \mathbb{Z}$) is a solution of $\sin \alpha = \frac{\sqrt{3}}{2}$ too and no other solution exists.
Now, in order to find your initial unknown $x$, all you have to do is to substitute back $3x=\alpha$ in (A) and solve for $x$. You’ll get:
$$\tag{B} x = 20^\circ + k 120^\circ \lor x = 40^\circ + k 120^\circ$$
A: Working in degrees,
$$2\sin 3x=\sqrt{3}\iff\sin 3x=\sin60^\circ=\sin120^\circ\\\iff\exists n\in\Bbb Z\left(3x=60^\circ+360n^\circ\lor3x=120^\circ+360n^\circ\right)\\\iff\exists n\in\Bbb Z\left(x=(20+120n)^\circ\lor x=(40+120n)^\circ\right).$$
