Decide all solutions to the following trigonometric equation: $\sin \left( 4\,x \right) =\cos \left( 3\,x \right).$ in the interval $[0, 2π[ $. Decide all solutions to the following trigonometric equation:
$$
\sin \left( 4\,x \right) =\cos \left( 3\,x \right).
$$
in the interval $[0, 2π[$. 
I start by expanding $\sin \left( 4\,x \right)$ and $\cos \left( 3\,x \right)$ to get to get that
$\sin \left( 4\,x \right) =4\,\sin \left( x \right) \cos \left( x
 \right)  \left( 1-2\, \left( \sin \left( x \right)  \right) ^{2}
 \right)$
$\cos \left( 3\,x \right) =\cos \left( x \right)  \left( 1-4\, \left( 
\sin \left( x \right)  \right) ^{2} \right)$.
I cancel $\cos \left( x \right)$ in both equations above and get that 
$4\,\sin \left( x \right)  \left( 1-2\, \left( \sin \left( x \right) 
 \right) ^{2} \right) =1-4\, \left( \sin \left( x \right)  \right) ^{2}$
which is equivalent to
$8\, \left( \sin \left( x \right)  \right) ^{3}-4\, \left( \sin \left( 
x \right)  \right) ^{2}-4\,\sin \left( x \right) +1=0$.
Now I use the substitution $u=\sin \left( x \right)$ to get to
$8\,{u}^{3}-4\,{u}^{2}-4\,u+1=0$ ⟺ $u \left( u-1 \right)  \left( 2\,u+1 \right) =-1/4$.
Now, I'm not so sure I'm on the right track. To solve the cubic equation for $u$, seem to lead to a real nasty piece of a solution. Appreciate any help!

Hint: Note that
$$
\sin(4x)=\cos(3x) = \sin\left(\frac{\pi}2-3x\right),
$$ and $\sin x = \sin y$ if and only if $x-y = 2n\pi$ or $x+y= (2n+1)\pi$ for some $n\in \Bbb Z$.
