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.

Note that by product to sum formula
$$2\cos \theta \sin \varphi = {{\sin(\theta + \varphi) - \sin(\theta - \varphi)} }\implies \cos 2x \cdot \sin x   = \frac{\sin 3x - \sin x}{2}
$$
