Number of solutions of $8\sin(x)=\frac{\sqrt{3}}{\cos(x)}+\frac{1}{\sin(x)}$ Number of solution in $[0,2\pi]$ satisfying the equation 
$$8\sin(x)=\frac{\sqrt{3}}{\cos(x)}+\frac{1}{\sin(x)}$$
Options are $5$ or $6$ or $7$ or $8$.
Doing some manipulations I reached $4\sin^2(x)\cos(x)=\sin(x+\pi/6)$ but I think this direction will not lead towards result. How to proceed?

Recall that $\sin x = \sqrt{1 - \cos^2 x}$
\begin{align}
8\sin^2x \cos x &= \sqrt{3} \sin x + \cos x\\
8(1 - \cos^2 x) &= \sqrt{3} \sqrt{1 - \cos^2 x} + \cos x
\end{align}
Let $a = \cos x$
\begin{align}
8(1 - a^2)a &= \sqrt{3}\sqrt{1 - a^2} + a\\
8(1 - a^2)a - a &= \sqrt{3}{\sqrt{1 - a^2}}
\end{align}
Then square both sides(this will not introduce extraneous solutions since $\sqrt{1 - \cos^2 x}$ is defined for all real $x$, since $-1 \leq \cos x \leq 1$).
\begin{align}
64a^2(1 - a^2)^2 - 16a^2(1 - a^2) + a^2 &= 3 - 3a^2\\
16a^2(1 - a^2)(4(1 - a^2) - 1) + 4a^2 - 3 &= 0\\
\end{align}
Let $a^2 = y$
\begin{align}
16y(1 - y)(4 - 4y - 1) + (4y - 3) &= 0\\
16y(1 - y)(3 - 4y) + (4y - 3) &= 0 \\
-16y(1 - y)(4y - 3) + (4y - 3) &= 0 \\
(1 - 16y(1 - y))(4y - 3) &= 0\\
(16y^2 - 16y + 1)(4y - 3) &= 0\\
\end{align}
The discriminant of the quadratic is positive, so we have 2 real solutions for the left factor, and one for the right factor. That yields 3 solutions for $y$. Recall that if $a^2 = y$ then $a = \pm \sqrt{y}$. This doubles the amount of solutions. Therefore there are 6 for $a$ solutions in the range $0 \leq x \leq 2\pi$. We can then use our substitution $a = \cos x$ to show that each of these solutions appears once in the given range. Henceforth there are 6 solutions to the equation in the desired range.
