Problem 6. Solve the equation $\arcsin \frac{x \sqrt{35}}{4 \sqrt{13}}+\arcsin \frac{x \sqrt{35}}{3 \sqrt{13}}=\arcsin \frac{x \sqrt{35}}{2 \sqrt{13}}$.
Answer: $0, \pm \frac{13}{12}$.

Solution. From the condition on the domain of the arcsine, it follows that

$$
|x| \leqslant \frac{2 \sqrt{13}}{\sqrt{35}}
$$

By calculating the sine of both sides of the equation and considering that $\cos \arcsin >0$, and therefore,

$$
\sin \arcsin A=A, \quad \cos \arcsin A=\sqrt{1-A^{2}}
$$

we obtain

$$
\frac{x \sqrt{35}}{4 \sqrt{13}} \sqrt{1-\frac{35 x^{2}}{9 \cdot 13}}+\frac{x \sqrt{35}}{3 \sqrt{13}} \sqrt{1-\frac{35 x^{2}}{16 \cdot 13}}=\frac{x \sqrt{35}}{2 \sqrt{13}}
$$

By moving everything to the left side of the equation, simplifying, and factoring out the common factor, we have

$$
\frac{x \sqrt{35}}{12 \cdot 13}\left(\sqrt{117-35 x^{2}}+\sqrt{208-35 x^{2}}-6 \sqrt{13}\right)=0
$$

From this equation, it follows that one solution of the equation is $x=0$, and the others are the roots of the equation

$$
\sqrt{117-35 x^{2}}+\sqrt{208-35 x^{2}}=6 \sqrt{13}
$$

From the condition on the domain of the arcsine, it follows that all radicands are positive. Since both sides of the equation are positive, they can be squared

$$
117-35 x^{2}+2 \sqrt{\left(117-35 x^{2}\right)\left(208-35 x^{2}\right)}+208-35 x^{2}=468
$$

By moving everything except the square root to the right side of the equation, we have

$$
2 \sqrt{\left(117-35 x^{2}\right)\left(208-35 x^{2}\right)}=70 x^{2}+143
$$

By squaring both sides of the equation again, we get

$$
4\left(117-35 x^{2}\right)\left(208-35 x^{2}\right)=4900 x^{4}+20020 x^{2}+20449
$$

or

$$
65520 x^{2}=76895
$$

Thus, the original equation has two more roots $x= \pm \frac{13}{12}$.

Verification. We verify that the left side of the equation lies in the interval $[-\pi / 2, \pi / 2]$ for the given values of the argument. For this, we calculate the cosine of the left side

$$
\begin{aligned}
\cos \left(\arcsin \frac{x \sqrt{35}}{4 \sqrt{13}}+\arcsin \frac{x \sqrt{35}}{3 \sqrt{13}}\right) & =\sqrt{\left(1-\frac{35 x^{2}}{16 \cdot 13}\right)\left(1-\frac{35 x^{2}}{9 \cdot 13}\right)}-\frac{35 x^{2}}{12 \cdot 13}= \\
& =\sqrt{\left(1-\frac{35 \cdot 13}{16 \cdot 144}\right)\left(1-\frac{35 \cdot 13}{9 \cdot 144}\right)}-\frac{35 \cdot 13}{12 \cdot 144}=\frac{43 \cdot 29}{1728}-\frac{455}{1728}>0
\end{aligned}
$$

Since the value of the cosine is positive, all the found numbers are solutions to the problem.