Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$ 
Solve
  $$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$

My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 x$
Solving and re-arranging we get 
$-64\sin^6 x+80\sin^4 x-20\sin^2 x+1=0$
Using the substitution $\sin^2 x=t$
$-64t^3+80t^2-20t+1=0$
I am not able to solve it from hence forth

Nice job so far. 
Notice that every exponent is even, so you can further substitute $t^2=y$, to obtain a cubic equation in $y$, which you can solve by the Cardano formula. 
