<s>[INST] Help with a trigonometry problem. I have encountered some problems to solve the left side to the right.

$$ \cos^ 2x \sin x   = \frac{\sin 3x + \sin x}{4}$$

I trying to solve a differential equation on the form $$ y'' + y = \cos^ 2x * \sin x$$ and need to rewrite it to  $$ y'' + y = \frac{\sin 3x + \sin x}{4}$$
Have tried with different combinations but not really got it to $$\frac{\sin 3x + \sin x}{4}$$
I have also tried to start backwards, but it seems not to be the method I would have chosen if I started from the left.
\begin{align}
   \frac{\sin 3x+\sin x}{4} 
 & =\frac{\sin x+\sin 3x}{4} \\ 
 & =\frac{\sin x+\sin \left( 2x+x \right)}{4} \\ 
 & =\frac{\sin x+\left( \sin 2x\cos x+\sin x\cos 2x \right)}{4} \\ 
 & =\frac{\sin x+\left( \left( 2\sin x\cos x \right)\cos x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\ 
 & =\frac{\sin x+\left( 2\sin x{{\cos }^{2}}x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\ 
 & =\frac{\sin x+\left( 2\sin x\left( 1-{{\sin }^{2}}x \right)+\sin x\left( 1-2{{\sin }^{2}}x \right) \right)}{4} \\ 
 & =\frac{\sin x+\left( (2\sin x-2{{\sin }^{3}}x)+(\sin x-2{{\sin }^{3}}x) \right)}{4} \\ 
 & =\frac{\sin x+\left( 3\sin x-4{{\sin }^{3}}x \right)}{4} \\ 
 & =\frac{4\sin x-4{{\sin }^{3}}x}{4} \\ 
 & =\frac{4\sin x(1-{{\sin }^{2}}x)}{4} \\ 
 & =\sin x{{\cos }^{2}}x \\ 
\end{align}
Would be grateful if you could give out some kind of guidance.
 [/INST] Consider
\begin{align}
\cos^2x\sin x&=(1-\sin^2x)\sin x\\
&=\sin x-\sin^3 x\\
&=\sin x-\frac14(3\sin x-\sin 3x) &(*)\\
&=\frac{\sin3x-\sin x}{4}
\end{align}

For $(*)$,
\begin{align}
\sin 3x&=\sin(x+2x)\\
&=\sin x\cos 2x+\sin 2x\cos x \\
&= \sin x (1-2\sin^2x)+2\sin x\cos^2x\\
&= \sin x-2\sin^3x+2\sin x(1-\sin^2x)\\
&=3\sin x-4\sin^3x
\end{align}
Hence,
$$\sin^3x=\frac14(3\sin x-\sin3x)$$
Basically, you will have to make some sort of detour around $\sin 3x$ and $\sin^3x$ no matter how you approach this problem...
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