My attempt at solving this:
$\sin(x) – \sqrt3\sin(3x) + \sin(5x) < 0$
$2\sin \left(\frac{5x+x}2\right) + \cos\left(\frac{5x-x}2\right) – \sqrt 3\sin(3x) < 0$
I divide everything with 2:
$\sin(3x) + \frac12\cos(2x) – \frac {\sqrt 3}2\sin(3x) < 0$
I think I have gone the wrong way at solving this problem. Please advise.
Best Answer
$$\sin (x)+\sin (5x) - \sqrt3\sin(3x) <0\Rightarrow 2\sin(3x)\cos (2x) - \sqrt{3}\sin(3x) < 0 \\\Rightarrow \sin(3x)\left(\cos(2x) -\frac{\sqrt{3}}{2}\right)<0 $$
Case $1$:
$$ \sin(3x) < 0 \text{ and } \cos (2x) > \frac{\sqrt3}2 \implies \frac{\pi}{12}<x<\frac\pi3$$
Case $2$:
$$ \sin (3x) > 0 \text{ and } \cos (2x) < \frac{\sqrt3}2 \implies \frac{2\pi}{3} <x< \frac{11\pi}{12}$$