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?

You have made some errors in your calculations (or some typos here).
$$\sin(3x)=\cos(2x)$$
$$ \sin(2x+x) = \cos(2x)$$
$$\sin(2x)\cos(x) + \cos(2x)sin(x) = \cos(2x) $$
$$ 2\sin(x)\cos(x)\cos(x) + (1-2\sin^2(x))\sin(x)) = \cos(2x) $$
$$ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^{\bf{3}}(x) = \cos(2x) $$
$$ 2\sin(x)(1-\sin^2(x))+\sin(x)-2\sin^{\bf{3}}(x)=\cos(2x) $$
$$ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^{\bf{3}}(x) = \cos(2x) $$
$$3\sin(x)-4\sin^3(x)=\cos(2x)$$
Then recall that $\cos(2x)=1-2\sin^2(x)$ to give:
$$3\sin(x)-4\sin^3(x)=1-2\sin^2(x)$$
This is a cubic in $\sin(x)$. For simplicity write $y=\sin(x)$ to get:
$$-4y^3+2y^2+3y-1=0$$
$$-(y-1)(4y^2+2y-1)=0$$
So $\sin(x)=1$ or $\sin(x)=\frac{-1\pm\sqrt{5}}{4}$
So $x=\frac{\pi}{2}$ or $x=\frac{\pi}{10}$, $\frac{9\pi}{10}$, $\frac{13\pi}{10}$, $\frac{17\pi}{10}$
