Solve $\sin(3x)=\cos(2x)$ 
Question: Solve $\sin(3x)=\cos(2x)$ for $0≤x≤2\pi$.

My knowledge on the subject; I know the general identities, compound angle formulas and double angle formulas so I can only apply those.
With that in mind
\begin{align} 
\cos(2x)=&~ \sin(3x)\\
\cos(2x)=&~ \sin(2x+x) \\
\cos(2x)=&~ \sin(2x)\cos(x) + \cos(2x)\sin(x)\\
\cos(2x)=&~  2\sin(x)\cos(x)\cos(x) + \big(1-2\sin^2(x)\big)\sin(x)\\
\cos(2x)=&~  2\sin(x)\cos^2(x) + \sin(x) - 2\sin^2(x)\\
\cos(2x)=&~  2\sin(x)\big(1-\sin^2(x)\big)+\sin(x)-2\sin^2(x)\\
\cos(2x)=&~  2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x)\\
\end{align}
edit 
\begin{gather}
 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x) = 1-2\sin^2(x) \\
 2\sin^3(x) - 3\sin(x) + 1 = 0 
\end{gather} 
This is a cubic right? 
So $u = \sin(x)$,
\begin{gather} 2u^3 - 3u + 1 = 0 \\ 
 (2u^2 + 2u - 1)(u-1) = 0 
\end{gather}
Am I on the right track?
This is where I am stuck what should I do now?

$$\cos2x=\sin3x=\cos\left(\dfrac\pi2-3x\right)$$
$$\iff2x=2m\pi\pm\left(\dfrac\pi2-3x\right)$$ where $m$ is any integer
Alternatively, $$\sin3x=\cos2x=\sin\left(\dfrac\pi2-2x\right)$$
$$3x=n\pi+(-1)^n\left(\dfrac\pi2-2x\right)$$  where $n$ is any integer
