Trigonometric functions - finding solutions Question:

Find the general solution for the equation:
  $$\sin x + 2\sin2x - \sin3x = 3$$

Approach:
Well using the identity of $\sin a - \sin b $, I merged together $\sin x - \sin3x$
And as $\sin 2x = 2\sin x\cos x$, I got up till here:
$$2\sin x(2\cos x - \cos 2x) = 3$$
Edit:

Considering $\sin x - \sin 3x = -2\cos 2x\sin x$
  The equation becomes:
  $$-2\cos 2x\sin x+2\sin 2x = 3$$
  $$(-2\sin x)\cos 2x +2\sin 2x = 3$$
  The maximum value of the equation $asinx + bcosx$ is $\sqrt{a^2 + b^2}$
  Using this in here: (considering $b = -2\sin x$ and $a = 2$)
  $$3 \leq \sqrt{4\sin^2 x + 4}$$
  $$3/2 \leq \sqrt{1 + sin^2x}$$
  Obviously cannot happen as maximum value of $\sqrt{1 + sin^2x}$ is $\sqrt2$


There are no real solutions.  $s = \sin(x)$ must satisfy
$16s^6-24s^3-12s^2+12s+9=0$, which has no real root.
