8.191. $\sin ^{-1} t-\sin ^{-1} 2 t=\sin ^{-1} 4 t$.

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8.191. $\sin ^{-1} t-\sin ^{-1} 2 t=\sin ^{-1} 4 t$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\sin t \neq 0, \\ \sin 2 t \neq 0, \\ \sin 4 t \neq 0 .\end{array}\right.$

Rewrite the equation as

$\frac{1}{\sin t}-\frac{1}{\sin 2 t}-\frac{1}{\sin 4 t}=0 \Rightarrow \sin 2 t \sin 4 t-\sin t \sin 4 t-\sin t \sin 2 t=0$.

Using the formula $\sin \alpha \sin \beta=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))$ we have

$\frac{1}{2}(\cos 2 t-\cos 6 t)-\frac{1}{2}(\cos 3 t-\cos 5 t)-\frac{1}{2}(\cos t-\cos 3 t)=0 \Leftrightarrow$

$\Leftrightarrow(\cos 2 t+\cos 5 t)-(\cos 6 t+\cos t)=0$.

Since $\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$, then

$$
2 \cos \frac{7 t}{2} \cos \frac{3 t}{2}-2 \cos \frac{7 t}{2} \cos \frac{5 t}{2}=0 \Leftrightarrow 2 \cos \frac{7 t}{2}\left(\cos \frac{3 t}{2}-\cos \frac{5 t}{2}\right)=0
$$

From this

1) $\cos \frac{7 t}{2}=0, \frac{7 t}{2}=\frac{\pi}{2}+\pi k, t_{1}=\frac{\pi}{7}+\frac{2}{7} \pi k, k \in Z$,
2) $\cos \frac{3 t}{2}-\cos \frac{5 t}{2}=0 \Leftrightarrow 2 \sin 4 t \sin t=0$ does not fit the domain of definition. Considering:

Domain of definition, $t=\frac{\pi}{7}+\frac{2}{7} \pi k=\frac{\pi}{7}(2 k+1), k \neq 7 l+3$, where $k$ and $l \in Z$.

Answer: $t=\frac{\pi}{7}(2 k+1), k \neq 7 l+3$, where $k$ and $l \in Z$.