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!

It'd be much easier to use the identities
$\sin \theta = \sin(2k \pi + \theta)$
$\cos \theta = \cos (\pm \theta +2k\pi)$
$\sin \theta = \cos (\frac \pi 2 - \theta); \cos(\theta) = \sin (\frac \pi 2-\theta)$.
So solutions occur when 
$4x = \pm (\frac \pi 2 - 3x) + 2k\pi$
$\pm 3x = \frac \pi 2 - 4x + 2k \pi$ so
$7x = \frac \pi 2 + 2k \pi$ or
$x = -\frac \pi 2 + 2k\pi$ or
$x = \frac \pi 2 + 2k \pi$ or
$x = \frac {(4k+1) \pi}{14}$ or $x= \frac {3\pi} 2$ or $x = \frac {\pi} 2$.
So $x = \frac 1{14}\pi, \frac 5{14}\pi, \frac 12 \pi, \frac 9{14} \pi, \frac {13}{14} \pi, \frac {17}{14}\pi, \frac 32\pi,$ or $ \frac {25}{14}\pi$
