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 $\sqrt2$


By using the trigonometric identities:
$$\sin(2x) = 2\sin(x)\cos(x)$$
$$\sin(3x) = 3\sin(x)-4\sin^{3}(x),$$ we have
$$\sin(x) +4\sin(x)\cos(x) -3\sin(x) + 4\sin^3(x) = 3 \implies \cos(x) = \frac{3 + 2\sin(x) -4\sin^3(x)}{4\sin(x)}.$$
By squaring the equation, the solutions are kept. Hence,
$$1-\sin^2(x) = \frac{16\sin^6(x)-16\sin^4(x)-24\sin^3(x)+4\sin^2(x)+12\sin(x) +9}{16\sin^2(x)}.$$
Take $y = \sin(x)$. Thus,
$$(1-y^2) = \frac{16y^6-16y^4-24y^3+4y^2 +12y+9}{16y^2}$$ $$\implies 16y^6-24y^3-12y^2 +12y+9 =0,$$ which has no real solutions.
