I know that $$4\cos^3{\theta}=\cos3{\theta}+3\cos{\theta}$$ and $$3\sin{\theta}=4\sin^3{\theta}+\sin3{\theta}$$
But that didn't help me to understand why $4\cos^3{\theta}=3\sin{\theta}$
EDIT: I was computing the value of $ \sin3{\theta}+\cos3{\theta}$ where $\sin{\theta}-\cos{\theta}=\frac{1}{2}$.
When I look at the answer it wrote:
$$(\sin{\theta}-\cos{\theta})^2=\frac{1}{4}=1-2\sin{\theta}\cos{\theta} \\{\therefore}\sin{\theta}\cos{\theta}=\frac{3}{8}\\\begin{align}\sin3{\theta}+\cos3{\theta}&=(-4\sin^3{\theta}+3\sin{\theta})+(4\cos^3{\theta}-3\cos{\theta})\\&=-4(\sin^3{\theta}-\cos^3{\theta})+3(\sin{\theta}-\cos{\theta})\end{align}$$
I thought it implies that $4\cos^3{\theta}=3\sin{\theta}$
Let's see if I understood what are you looking for. Problem: Given that $\sin x - \cos x =\frac12$, find $\sin(3x) +\cos(3x)$. Is that it?
From what you said: $$\cos(3x)=4\cos^3 x-3\cos x$$ $$\sin(3x)=3\sin x -4\sin^3x$$ Add those up to get: $$\sin(3x)+\cos(3x)=3(\sin x- \cos x)-4(\sin^3x-\cos^3x)$$ You also nicely computed $$\sin x \cdot\cos x=\frac38$$ And since: $$\sin^3x-\cos^3x= (\sin x-\cos x)(1+\sin x\cos x)$$ We get that $$\sin(3x)+\cos(3x)=3\cdot\frac12 -4 \left(\frac12\left(1+\frac38\right)\right)$$