<s>[INST] 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?
 [/INST] Use $\sin 3x=3 \sin x - 4 \sin^3x$ and $\cos 2x=1-2\sin^2x$. To get
$$3 \sin x - 4 \sin^3x=1-2\sin^2x.$$
Now call $\sin x=t$. Thus we have
$$4t^3-2t^2-3t+1=0.$$
Observe that $t=1$ is definitely a solution, so we have
$$(t-1)(4t^2+2t-1)=0.$$ 
The quadratic factor will be zero for 
$$t=\frac{-1\pm \sqrt{5}}{4}$$
I hope you can solve from here. 
</s>