<s>[INST] 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?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
 [/INST] Ok, there might be nicer ways to do this but here it goes. Let $\theta=\pi/15$ so you have
$$
\tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta)
$$
then you use $\tan()=\frac{\sin()}{\cos()}$, and then you use the equivalences
$2\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)-\cos(\alpha+\beta),$
and
$2\cos(\alpha)\cos(\beta)=\cos(\alpha-\beta)+\cos(\alpha+\beta)$.
You should get
$\frac{\cos(3\theta)\cos(\frac{3}{2}\theta)-\cos(3\theta)\cos(\frac{21}{2}\theta)-\cos(5\theta)\cos(\frac{3}{2}\theta)+\cos(5\theta)\cos(\frac{21}{2}\theta)}{\cos(3\theta)\cos(\frac{3}{2}\theta)+\cos(3\theta)\cos(\frac{21}{2}\theta)+\cos(5\theta)\cos(\frac{3}{2}\theta)+\cos(5\theta)\cos(\frac{21}{2}\theta)} $
** I also used the fact that $\cos(-\alpha)=\cos(\alpha)$. Now, note that you have something of the form
$\frac{a-b-c+d}{a+b+c+d},$
which is equal to $1+\frac{-2b-2c}{a+b+c+d}$. Now, we shall show that $-2b-2c=0$. So we have
$
-2b-2c=-2\cos(3\theta)\cos(\frac{21}{2}\theta)-2\cos(5\theta)\cos(\frac{3}{2}\theta)
$
using the formula above for the product of cosines we get
$-2\cos(3\theta)\cos(\frac{21}{2}\theta)-2\cos(5\theta)\cos(\frac{3}{2}\theta)=-\cos(\frac{15}{2}\theta)-\cos(\frac{27}{2}\theta)-\cos(\frac{7}{2}\theta)-\cos(\frac{13}{2}\theta).$
Now plug in the value of $\theta$, you get
$
-\cos(\pi/2)-\cos(\frac{27}{30}\pi)-\cos(\frac{7}{30}\pi)-\cos(\frac{13}{30}\pi)
$
naturally the first element is $0$. Now, using the fact that $\cos(\alpha\pm\pi/2)=\mp\sin(\alpha)$ you can write
$
-\cos(\frac{27}{30}\pi)-\cos(\frac{7}{30}\pi)-\cos(\frac{13}{30}\pi)=\sin(6\pi/15)-\sin(\frac{4}{15}\pi)-\sin(\pi/15)
$
Now you use $\sin(\alpha)\cos(\beta)=\frac{\sin(\alpha+\beta)+\cos(\alpha-\beta)}{2}$ looking for two numbers such that the equality holds. You should get then
$
\sin(6\pi/15)-\sin(\frac{4}{15}\pi)-\sin(\pi/15)=\sin(6\pi/15)-2\sin(\pi/6)\cos(\pi/10)=\sin(6\pi/15)-\cos(\pi/10)
$
which it is easily seen to be zero.
Therefore
$
\tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta)=1, \qquad \theta=\pi/15
$
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