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.
 A: \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 $\, 162^{\circ}$

A: Using the trigometric identities you can write 
$$\sin 2\theta=2\sin\theta\cos\theta$$
$$\cos2\theta=\cos^2\theta-\sin^2\theta$$
$$\cos3\theta=\cos^3\theta-\sin^2\theta\cos\theta$$
$$\sin4\theta=4\sin\theta\cos\theta\left(\cos2\theta-\sin^2\theta\right)$$
Then you get
$$2\sin\theta\cos\theta\left(1+2\cos^2\theta-2\sin^2\theta\right)=\cos\theta\left(1+\cos^2\theta-\sin^2\theta\right)$$
This means $\cos\theta=0$ ($\theta=\pi/2$) is at least one solution.
A: Using
$$\sin x-\cos x =\sqrt{2} \sin\left(x-\frac{\pi}{4}\right),$$
given equation can be represented as
$$\cos \theta \sin \left(3\theta - \frac{\pi}{4}\right)=0.$$
A: It becomes
$$\sin3\theta\, \cos\theta=\cos2\theta\, \cos\theta, $$
First the common factor solution
$$ \cos \theta = 0 \rightarrow \theta = ( 2 n - 1 ) \pi/2 $$
First solution $$ \pi/2$$
Next solve what remains
$$ \sin 3\theta = = \cos 2 \theta = \sin (\pi/2- 2 \theta) $$
$$ 5 \theta = \pi/2 $$
$$ \theta_1 = \pi/10 ,\theta_2 = \pi-\pi/10  = 9 \pi/10. $$
A: Using Prosthaphaeresis Formulas as suggested,
$$ \sin 2\theta+\sin 4\theta = \cos \theta+\cos 3\theta \\
  2\sin 3\theta \cos \theta = 2\cos 2\theta \cos \theta \\
  \cos \theta \, (\sin 3\theta-\cos 2\theta) = 0 $$
If $\cos \theta=0,\theta=(2n+1)90^\circ$ where $n$ is any integer
We need $0<(2n+1)90^\circ<180^\circ\implies-1<n<1\implies n=0$
Else $\cos 2\theta=\sin 3\theta=\cos(90^\circ-3\theta)$
$2\theta=360^\circ m\pm(90^\circ-3\theta)$
$+\implies\theta=72^\circ m+18^\circ$ and we need $0<72^\circ m+18^\circ<180^\circ\implies m=?$
$-\implies\theta=-360^\circ m+90^\circ$
Can you take it from here?
