Solving $\sin x\cos3x+\cos x\sin3x=\sqrt{3}/2$ in the interval $[0,2\pi]$ 
Solve the following trig equation:
$$\sin(x)\cos(3x)+\cos(x)\sin(3x)=\frac{\sqrt{3}}{2}$$
in the interval $[0,2\pi]$.

Using trig identities, it is now at
$$\sin(x+3x)=\sqrt{3}/2 \tag{1}$$
Next, it has become
$$\sin(x+3x)=\sin 60^\circ \tag{2}$$
leading to
$$\sin(x+3x)-\sin60^\circ=0 \tag{3}$$
From there, it is unclear what to do.

$\sin(\theta)=\frac{\sqrt{3}}{2}$ when $\theta=\frac{\pi}{3}$, as you pointed out, but also when $\theta=\frac{2\pi}{3}$. Not only that, because the sine function is periodic, we must also remember that adding any multiple of $2\pi$ to either of those angles gives another solution to $\sin(\theta)=\frac{\sqrt{3}}{2}$.
In general we have $\theta=\frac{\pi}{3}+2\pi\cdot k$ and $\theta=\frac{2\pi}{3}+2\pi\cdot k$, where $k$ can be any integer. Note that I am using radians because the answer needs to be in radians in the interval $[0,2\pi]$.
Since $\theta$ represents $4x$, we need to solve the two equations above, substituting $4x$ in place of $\theta$.
For the first, we have $4x=\frac{\pi}{3}+2\pi\cdot k$
Divide by $4$ to get $x=\frac{\pi}{12}+\frac{\pi}{2}\cdot k$.
Now we consider what values of $k$ will make $x$ be in the interval $[0,2\pi]$
We can see that $k= 0, 1, 2, 3$, corresponds to $x=\frac{\pi}{12}, \frac{7\pi}{12}, \frac{13\pi}{12}, \frac{19\pi}{12}$
I will leave solving $4x=\frac{2\pi}{3}+2\pi\cdot k$ up to you
