Solving Trigonometric Equation. 
Solve for $\theta$ $[0°<\theta<180°]$
  $$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$

My solution is here:
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
After using the transformation formula, I got
$$\sin3\theta\cos\theta=\cos3\theta\cos\theta.$$
I could not proceed from here.

\begin{array}{rcl}
  \sin 2\theta+\sin 4\theta &=& \cos \theta+\cos 3\theta \\
  2\sin 3\theta \cos \theta &=& 2\cos 2\theta \cos \theta \\
  \cos \theta \, (\sin 3\theta-\cos 2\theta) &=& 0 \\
  \cos \theta \, (3\sin \theta-4\sin^{3} \theta+2\sin^{2} \theta-1)
  &=& 0 \\
  \cos \theta \, (1-\sin \theta)(4\sin^{2} \theta+2\sin \theta-1) &=& 0 \\
\end{array}
$\therefore \, \cos \theta=0$, $\, \sin \theta=1 \,$, or 
$\displaystyle \, \sin \theta=\frac{-1\pm \sqrt{5}}{4}$
$\because \, 0 < \theta < 180^{\circ}$, reject $\, \sin \theta \leq 0$
$\therefore \, \theta = 18^{\circ}$, $90^{\circ}$ (twice) or $\, 162^{\circ}$

