Analytically solving $\frac{1}{\sin2x} + \frac{1}{\sin3x} = \frac{1}{\sin x}$

Question

Given $$ \frac{1}{\sin(2x)} + \frac{1}{\sin(3x)} = \frac{1}{\sin x}$$

I tried solving the equation above using the double and triple angle formulas and arrived at this cubic expression in $\cos x$

$$ 8\cos^3(x)-4 \cos^2(x)-4\cos(x) + 1$$

I ( and apparently wolfram alpha too) and unable to solve it analytically. But I when I take the inverse cosine of the “numerical” roots, i get exact answers, namely $\frac{\pi}{7}$, $\frac{5\pi}{7}$, and $\frac{3\pi}{7}$. How should I approach problems like these?

Answer 1

Hint:

Remember $\sin x\ne0\implies\cos x\ne\pm1$

From $$(2c)^3-(2c)^2-2(2c)+1=0$$

Replace $2c$ with $y+\dfrac1y$ where $y=e^{ix}$ to find

$$\left(y+\dfrac1y\right)^3-\left(y+\dfrac1y\right)^2-2\left(y+\dfrac1y\right)+1$$ $$=y^3+\dfrac1{y^3}+3\left(y+\dfrac1y\right)-y^2-\dfrac1{y^2}-2-2\left(y+\dfrac1y\right)+1$$ $$=\dfrac{y^7-1}{y^3(y-1)}$$

Answer 2

First observe that $\sin x\ne0\implies\cos x\ne\pm1$

If I start calculation from $$\dfrac1{\sin2x}+\dfrac1{\sin3x}=\dfrac1{\sin x}$$

$$2\sin x(\sin2x+\sin3x)=2\sin2x\sin3x$$

Using Werner formula and on simplification,

$$\iff\cos5x+\cos2x=\cos4x+\cos3x$$

Apply Prosthaphaeresis Formulas as $5+2=4+3$