<s>[INST] Analytically solving $\frac{1}{\sin2x} + \frac{1}{\sin3x} = \frac{1}{\sin x}$ 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?
 [/INST] 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
$$\left(y+\dfrac1y\right)^3-\left(y+\dfrac1y\right)^2-2\left(y+\dfrac1y\right)+1$$
$$=y^3+\dfrac1{y^3}+3\left(y+\dfrac1y\right)-y^2-\dfrac1{y^2}-2-2\left(y+\dfrac1y\right)+1$$
$$=\dfrac{y^7-1}{y^3(y-1)}$$ 
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