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

Question

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

Answer 1

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 $\sin^2x$

Answer 2

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.

Answer 3

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

Answer 4

We need to solve $$8\sin^2x\cos{x}=\sqrt3\sin{x}+\cos{x}$$ or $$2\sin2x\sin{x}=\cos(x-60^{\circ})$$ or $$\cos{x}-\cos3x=\cos(x-60^{\circ})$$ or $$2\sin30^{\circ}\sin(30^{\circ}-x)=\cos3x$$ or $$\cos(60^{\circ}+x)=\cos3x.$$ Can you end it now?

Answer 5

Hint to solve the equation you got

If you make $t=\sin^{2}x$ 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 $\sin^2x=\dfrac1z.$