<s>[INST] 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
 [/INST] First thing first, if you make the substitution $t=\sin^2x$ the polynomial you get is $$-64t^3+80t^2-20t+1=0$$ Now, to decompose it, you could use Ruffini's rule: first we find a zero of the polynomial that divides the constant term (in our case $\pm 1$). Let us call the polynomial $P(t)$, then $$P(1) =-64+80-20+1 \neq 0 \\P(-1) = 64+80+20+1 \neq 0$$ clearly we have no integer solutions! So what we can do is substitute $z=\frac{1}{t}$ and what we get is the following polynomial in $z$ $$Q(z) =   z^3-20z^2+80z-64$$ and now we apply the same rule: let us find a zero of $Q(z)$ in the divisors of the constant term $64$ which are $\pm1,\pm2,\pm4,\cdots$. You can easily see that $$Q(4) = 0$$ then we can go on with the simplification and get $$(z-4)(z^2-16z+16)=0$$ which is definetly easier to solve. We find $$z_1=4\implies t_1={1\over 4}\\ z_2=4(2-\sqrt{3})\implies t_2 = {1\over {4(2-\sqrt{3})}}\\ z_3 = 2(2+\sqrt{3})\implies t_3 = {1\over {4(2+\sqrt{3})}}$$ from which you can find the values of $\sin^2x$
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