Solve the following equation:

$$
\frac{1}{\sin x}-\frac{1}{\sin 2 x}=\frac{2}{\sin 4 x}
$$
Solution. Transform the denominators using the known angle sum formulas:

$$
\begin{aligned}
& \sin 2 x=2 \sin x \cos x \\
& \sin 4 x=\sin (2 x+2 x)=4 \sin x \cos x\left(\cos ^{2} x-\sin ^{2} x\right)=4 \sin x \cos x\left(2 \cos ^{2} x-1\right)
\end{aligned}
$$

From the domain of definition, we must exclude those values where the denominator is zero: $\sin x=0, \cos x=0$, and $\cos x= \pm \sqrt{\frac{1}{2}}$

Bring to a common denominator. The common denominator is: $4 \sin x \cos x\left(2 \cos ^{2} x-1\right)$. Multiplying by this, we get the following equation:

$$
4 \cos x\left(2 \cos ^{2} x-1\right)-2\left(2 \cos ^{2} x-1\right)=2
$$

After rearranging, we get

$$
2 \cos ^{3} x-\cos ^{2} x-\cos x=0
$$

From this,

$$
\cos x\left(2 \cos ^{2} x-\cos x-1\right)=0
$$

We have excluded the possibility $\cos x=0$. From the equation $2 \cos ^{2} x-\cos x-1=0$, $\cos x=1$ is not possible because then $\sin x=0$ would be, but we have excluded this. The other root is $\cos x=-\frac{1}{2}$, and from this

$$
x= \pm \frac{2 \pi}{3}+2 n \pi, \quad \text { where } \quad n \in \mathbb{Z}
$$