<s>[INST] Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ 
For what values of $x$ between $0$ and $\pi$ does the inequality $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ hold?

My Attempt
$$
\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\
x\in(0,\pi)\implies4x\in(0,4\pi)\\
4x\in(0,\pi)\cup(2\pi,3\pi)\implies x\in\Big(0,\frac{\pi}{4}\Big)\cup\Big(\frac{\pi}{2},\frac{3\pi}{4}\Big)
$$
But, my reference gives the solution, $x\in\Big(0,\dfrac{\pi}{4}\Big)\cup\Big(\dfrac{3\pi}{4},\pi\Big)$, where am I going wrong with my attempt?
 [/INST] The given solution is wrong; you are correct. At $x=\frac{7\pi}8\in\left(\frac{3\pi}4,\pi\right)$, we have that $$\frac14\sin4x=\frac14\sin\frac{7\pi}2=-\frac14<0$$ which is a contradiction.
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