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.
On
A pretty straightforward question, as you have used the identity, $\sin(A+B)=\sin A \cos B + \cos A \sin B$
You get, $\sin(4x)=\frac{\sqrt{3}}{2}$
Either you can make the graph of $\sin(4x)$ (having a period of $\frac{2\pi}{4}$) or make a unit circle, however you prefer, and see where it crosses $\frac{\sqrt{3}}{2}$
Or you can stick to the general solution ,
for sine it is, if $\sin(\theta)=sin(\phi) \implies \theta = n\pi + (-1)^n\phi, n \in \mathbb{Z}$,
here $\theta$ is $4x$
So, $x=\frac{n\pi}{4}+\frac{(-1)^n\pi}{12}$
Substitute values of $n$ for which $x \in [0,2\pi]$ and that will the whole set of solutions of the given equation.
$\sin(\theta)=\frac{\sqrt{3}}{2}$ 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