Problem 6. Solve the equation $\arcsin \frac{x \sqrt{5}}{3}+\arcsin \frac{x \sqrt{5}}{6}=\arcsin \frac{7 x \sqrt{5}}{18}$.
Answer: $0, \pm \frac{8}{7}$.

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

$$
|x| \leqslant \frac{18}{7 \sqrt{5}}
$$

By computing 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

$$
\frac{x \sqrt{5}}{3} \sqrt{1-\frac{5 x^{2}}{36}}+\frac{x \sqrt{5}}{6} \sqrt{1-\frac{5 x^{2}}{9}}=\frac{7 x \sqrt{5}}{18}
$$

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

$$
\frac{x \sqrt{5}}{18}\left(\sqrt{36-5 x^{2}}+\sqrt{9-5 x^{2}}-7\right)=0
$$

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

$$
\sqrt{36-5 x^{2}}+\sqrt{9-5 x^{2}}=7
$$

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

$$
36-5 x^{2}+2 \sqrt{\left(36-5 x^{2}\right)\left(9-5 x^{2}\right)}+9-5 x^{2}=49
$$

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

$$
2 \sqrt{\left(36-5 x^{2}\right)\left(9-5 x^{2}\right)}=10 x^{2}+4
$$

By squaring both sides of the equation again, we get

$$
4\left(36-5 x^{2}\right)\left(9-5 x^{2}\right)=100 x^{4}+80 x^{2}+16
$$

or

$$
980 x^{2}=1280
$$

Thus, the original equation has two more roots $x= \pm \frac{8}{7}$.

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 compute the cosine of the left side

$$
\begin{aligned}
\cos \left(\arcsin \frac{x \sqrt{5}}{3}+\arcsin \frac{x \sqrt{5}}{6}\right)= & \sqrt{\left(1-\frac{5 x^{2}}{9}\right)\left(1-\frac{5 x^{2}}{36}\right)}-\frac{5 x^{2}}{18}= \\
& =\sqrt{\left(1-\frac{5 \cdot 64}{9 \cdot 49}\right)\left(1-\frac{5 \cdot 64}{36 \cdot 49}\right)}-\frac{5 \cdot 64}{18 \cdot 49}=\frac{11 \cdot 38}{800}-\frac{320}{882}>0
\end{aligned}
$$

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