Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$

Let's use degree measure (for writing convenience):
we need to prove identity
$$
\tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ = 1.\tag{1}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \cos 12^\circ \cos 48^\circ \cos 54^\circ \cos 72^\circ;\tag{2}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \sin 78^\circ \sin 42^\circ \sin 36^\circ \sin 18^\circ;
$$
$$
(\sin 12^\circ \sin 72^\circ ) \cdot ( \sin 48^\circ \sin 54^\circ) =^?= (\sin 18^\circ \sin 78^\circ ) \cdot ( \sin 36^\circ \sin 42^\circ);
$$
$$
(\cos 60^\circ - \cos 84^\circ)(\cos 6^\circ - \cos 102^\circ) =^?=
(\cos 60^\circ - \cos 96^\circ)(\cos 6^\circ - \cos 78^\circ);
$$
$$
(\cos 60^\circ - \cos 84^\circ)(\cos 6^\circ + \cos 78^\circ) =^?=
(\cos 60^\circ + \cos 84^\circ)(\cos 6^\circ - \cos 78^\circ);
$$
$$
-\cos 84^\circ \cos 6^\circ + \cos 60^\circ \cos 78^\circ =^?=
\cos 84^\circ \cos 6^\circ - \cos 60^\circ \cos 78^\circ;\tag{3}
$$
$$
\cos 84^\circ \cos 6^\circ - \cos 60^\circ \cos 78^\circ=^?=0.\tag{4}
$$
Yes, $(4)$ is true  identity, because
$$
2\sin 6^\circ \cos 6^\circ = \sin 12^\circ,
$$
$$
\sin 6^\circ \cos 6^\circ = \frac{1}{2}\sin 12^\circ,
$$
$$
\cos 84^\circ \cos 6^\circ = \cos 60^\circ \cos 78^\circ.\tag{5}
$$
So, we get equation
$$
\sin (x) = 1. \tag{6}
$$
Solution is obvious...
