<s>[INST] 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 $\sqrt2$

 [/INST] By setting $z=e^{ix}\in S^1$, the given equation can be written as:
$$ \Im\left(z+2z^2-z^3\right) = 3 \tag{1}$$
that implies:
$$ \left\|z+2z^2-z^3\right\| = \left\| 2+\frac{1}{z}-z \right\|\geq 3\tag{2}$$
but that cannot happen since:
$$ \left\| 2+\frac{1}{z}-z \right\| = \left\| 2-2i\sin x\right\|\leq 2\sqrt{2} < 3,\tag{3} $$
so there are no real solutions.
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