To solve $\sin x= 2\sin(x-30^\circ) \sin(40^{\circ})$ I expanded RHS by using $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$,
$$\sin x= 2\sin(40^{\circ})\times\left(\frac{\sqrt3}2\sin x- \frac12\cos x\right)$$
$$\left(\sqrt3 \sin(40^{\circ})-1\right)\sin x=
\sin(40^{\circ})\cos x$$
$$\cot x= \sqrt 3- \csc(40^{\circ})$$
But from here I'm not sure how to continue.
I've also tried using $2\sin\alpha\sin\beta= \cos(\alpha-\beta)-\cos(\alpha+\beta)$ for the RHS which result in, $\sin x= \cos(x-70^\circ)-\cos(x+10^\circ)$ but don't know how to continue.
Best Answer
Since $\sin 40^\circ = \cos (90^\circ - 40^\circ) = \cos 50^\circ$, we have $$\begin{align} 0 &= -\sin x + 2 \sin (x - 30^\circ) \cos 50^\circ \\ &= -\sin x + \sin (x + 20^\circ) + \sin (x - 80^\circ) \\ &= 2 \sin 10^\circ \cos (x + 10^\circ) + \sin (x - 80^\circ) \\ &= 2 \sin 10^\circ \sin (90^\circ - (x + 10^\circ)) + \sin (x - 80^\circ) \\ &= 2 \sin 10^\circ \sin (80^\circ - x) + \sin (x - 80^\circ) \\ &= (1 - 2 \sin 10^\circ) \sin (x - 80^\circ). \end{align}$$ Since $1 - 2 \sin 10^\circ \ne 0$, this immediately implies $$x = 80^\circ + 180^\circ k$$ for any integer $k$.