Trigonometric equation: $3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$ 
Solve the following equation:
$$3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$$

My attempt:
$$3\sin x = -\cot x \cdot \cot 2x \cdot (\tan^2 x + \tan 2x)$$
$$\implies 3\sin x = -\tan x \cdot \cot 2x - \cot x$$
$$\implies 3\sin x + \tan x \cdot \cot 2x + \cot x = 0$$
Since $\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x} \implies \cot 2x = \dfrac{1}{\tan 2x} = \dfrac{1 - \tan^2 x}{2 \tan x}$, so:
$$3\sin x + \dfrac{1 - \tan^2 x}{2} + \cot x = 0$$
I got stuck from here. How can I proceed this?
 A: I think that after writing that tou want to find the zeros of
$$f(x)=3\sin (x) + \dfrac{1 - \tan^2 (x)}{2} + \cot (x) $$ the problem becomes purely numerical.
If you plot the function, it is quite awful because of the discontinuities due to the tangent and cotangent. It would be better to remove them multipluing everythng by $\sin(x)$ (because of the cotangent) and by $\cos^2(x)$ (because of the tangent). Doing it and using a few trigonometric identities, you could instead try to find the zeros of function
$$g(x)=3-2 \sin (x)+2 \sin (3 x)+6 \cos (x)+2 \cos (3 x)-3 \cos (4 x)$$
Ploting the funtion $g(x)$, we see positive solutions close to $1.3$, $2.0$, $2.6$, $3.8$, $7.5$, $10.0$, $\dots$ and negative solutions close to $-11.3$, $-10.6$, $-10.0$, $-8.8$, $-2.5$.
Using any of these as a starting point, using Newton method will converge quite fast. For example for the fourth positive root, the iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 3.80000 \\
 1 & 3.74477 \\
 2 & 3.74671 \\
 3 & 3.74671
\end{array}
\right)$$ As said in comments, there is more than likely a problem with the problem itself.
As @am301 commented, making $y=\sin(x)$ leads to a monster.
Using the tangent half-angle substitution would require to solve for $t$
$$\frac {t^8-t^7-14 t^6+5 t^5+24 t^4+5 t^3-10 t^2-t-1 } {2 t\left(t^2-1\right)^2 \left(t^2+1\right) }=0$$ the real solutions of which being
$$\{ -3.20369, 0.69728, 1.50324, 3.8625\}$$
