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?
 A: 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.
A: Finishing what you did
$$\sin(4x)>0\iff$$
$$2k\pi <4x<(2k+1)\pi \iff$$
$$\frac{k\pi}{2}<x<\frac{k\pi}{2}+\frac{\pi}{4}$$
$k=0$ gives $0<x<\frac{\pi}{4}$
and
$k=1$ gives $\frac{\pi}{2}<x<\frac{3\pi}{4}$.
A: As an alternative for a full solution we can consider two cases


*

*$\sin x \cos x >0$ that is $x\in(0,\pi/2)\cup(\pi,3\pi/2)$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x>\sin^2x \iff2\sin^2 x<1$$
$$-\frac{\sqrt 2}2<\sin x<0 \,\land\, 0<\sin x<\frac{\sqrt 2}2 \iff \color{red}{x\in(0,\pi/4)}\cup(\pi,5\pi/4)$$


*

*$\sin x \cos x <0$ that is $x\in(\pi/2,\pi)\cup(3\pi/2,2\pi)$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x<\sin^2x \iff2\sin^2 x>1$$
$$-1<\sin x<-\frac{\sqrt 2}2\,\land\, \frac{\sqrt 2}2<\sin x <1 \iff \color{red}{x\in(\pi/2,3\pi/4)}\cup(3\pi/2,7\pi/4)$$
and then your solution is correct.
