Vtyjkyuk

Solve
$$8sin x=dfracsqrt3cos x+frac1sin x$$

My approach is as follow
$8 sin x-frac1sin x=fracsqrt3cos x$

On squaring we get

$64 sin^2 x+frac1sin^2 x-16=frac3cos^2 x$

$(64sin^4 x-16sin^2 x+1)(1-sin^2 x)=3 sin^2 x$

Solving and re-arranging we get
$-64sin^6 x+80sin^4 x-20sin^2 x+1=0$

Using the substitution $sin^2 x=t$

$-64t^3+80t^2-20t+1=0$

I am not able to solve it from hence forth

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 $z=frac1t$ 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 $$z_1=4implies t_1=1over 4\ z_2=4(2-sqrt3)implies t_2 = 1over 4(2-sqrt3)\ z_3 = 2(2+sqrt3)implies t_3 = 1over 4(2+sqrt3)$$ from which you can find the values of $sin^2x$

We need to solve
$$8sin^2xcosx=sqrt3sinx+cosx$$ or
$$2sin2xsinx=cos(x-60^circ)$$ or
$$cosx-cos3x=cos(x-60^circ)$$ or
$$2sin30^circsin(30^circ-x)=cos3x$$ or
$$cos(60^circ+x)=cos3x.$$
Can you end it now?

$$−64t^6+80t^4−20t^2+1=(1-4t^2)(1-16t^2+16t^4)$$

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.

Hint to solve the equation you got

If you make $t=sin^2x$ then you get $-64t^3+80t^2-20t+1=0.$ It has no integer solutions. So we consider $z=1/t.$ Then we have

$$z^3-20z^2+80z-64=0.$$ It easy to see that $4$ is a root. So we have

$$z^3-20z^2+80z-64=(z-4)(z^2-16z+16).$$ Solve the quadratic equation and finally use that $sin^2x=dfrac1z.$