Solve $\sin(x)-\sin(\frac{\pi}{3}-x)=\sqrt{\frac{3}{2}}$ As in the title, solve $\sin(x)-\sin(\frac{\pi}{3}-x)=\sqrt{\frac{3}{2}}$. I was trying to rewrite it into a simpler form, but without any luck.
 A: Note that
$$\sin(x)-\sin(\frac\pi3-x) = \sin(x)+\sin(x-\frac\pi3) = \sqrt3 \sin(x-\frac\pi6) $$
(draw a sketch for the latter identity, using the parallelogram rule to add the two sines).
So now you want to solve $\sin(x-\pi/6) = 1/\sqrt2$, which gives $$x-\frac\pi6 = (2\pm 1)\frac\pi4$$
A: $$\sin(x)-\sin\left(\frac{\pi}{3}-x\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$\sin(x)-\cos\left(\frac{\pi}{6}+x\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$\frac{3\sin(x)}{2}-\frac{1}{2}\sqrt{3}\cos(x)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$-\sqrt{3}\left(\frac{\cos(x)}{2}-\frac{1}{2}\sqrt{3}\sin(x)\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$-\sqrt{3}\left(\cos\left(\frac{\pi}{3}\right)\cos(x)-\sin\left(\frac{\pi}{3}\right)\sin(x)\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$-\sqrt{3}\sin\left(\frac{\pi}{6}-x\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{6}-x\right)=-\frac{1}{\sqrt{2}}\Longleftrightarrow$$
$$\frac{\pi}{6}-x=\frac{5\pi}{4}+2\pi n_1\Longleftrightarrow\space\space\vee\space\space\frac{\pi}{6}-x=\frac{7\pi}{4}+2\pi n_2\Longleftrightarrow$$
$$-x=\frac{13\pi}{12}+2\pi n_1\Longleftrightarrow\space\space\vee\space\space -x=\frac{19\pi}{12}+2\pi n_2\Longleftrightarrow$$
$$x=-\frac{13\pi}{12}-2\pi n_1\space\space\vee\space\space x=-\frac{19\pi}{12}+2\pi n_2$$
With $n_1,n_2\in\mathbb{Z}$
