Problem 3. Solve the equation

$$
\arcsin 2 x+\arcsin x=\frac{\pi}{3}
$$

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The first solution. Since the function $\arcsin x$ is defined on the interval $[-1 ; 1]$, the function $\arcsin 2 x$ is defined on the interval $[-0.5 ; 0.5]$. Therefore, the monotonic (increasing) and continuous function $f(x) = \arcsin 2 x + \arcsin x$ is defined on the interval $[-0.5 ; 0.5]$. We find:

$$
\begin{aligned}
f(-0.5) & = \arcsin (-1) + \arcsin (-0.5) = \\
& = -0.5 \pi - \frac{\pi}{6} = -\frac{2}{3} \pi
\end{aligned}
$$

$$
f(0.5) = \arcsin 1 + \arcsin 0.5 = 0.5 \pi + \frac{\pi}{6} = \frac{2}{3} \pi
$$

Now it is clear that equation (1) has a unique root $x_{1}$, and $0 < x_{1} < 0.5$.

Using a calculator, we find the approximate value of $x_{1}$ with an accuracy of $10^{-6}$. For this, we create the following table of values of the function $f(x)$. Note that $\frac{\pi}{3} \approx 1.0471975$:

| $x$ | $f(x)$ | $x$ | $f(x)$ |
| :--- | :--- | :--- | :--- |
| 0 | 0 | 0.3279 | 1.0470982 |
| 0.5 | 2.0943951 | 0.32735 | 1.0472834 |
| 0.3 | 0.9481938 | 0.32733 | 1.0472093 |
| 0.35 | 1.129686 | 0.32732 | 1.0471723 |
| 0.33 | 1.0571224 | 0.327325 | 1.0471908 |
| 0.328 | 1.0496925 | 0.327327 | 1.0471982 |
| 0.327 | 1.0459873 | 0.3273265 | 1.0471963 |
| 0.3275 | 1.0478391 |  |  |

Answer. $x_{1} \approx 0.3273227$.

The second solution. We find the cosine of both sides of equation (1):

$$
\cos (\arcsin 2 x + \arcsin x) = \cos \frac{\pi}{3}
$$

From this,

$$
\sqrt{1-4 x^{2}} \cdot \sqrt{1-x^{2}} - 2 x \cdot x = 0.5
$$

Solving this irrational equation, we get:

$$
x_{1} = 0.5 \sqrt{\frac{3}{7}} ; x_{2} = -0.5 \sqrt{\frac{3}{7}} .
$$

The obtained solution needs verification. Let $\alpha = \arcsin 2 x_{1} + \arcsin x_{1}$. Then $\cos \left(\arcsin \sqrt{\frac{3}{7}} + \arcsin \left(0.5 \sqrt{\frac{3}{7}}\right)\right) = \cos \alpha$ and $\cos \alpha = \sqrt{1 - \frac{3}{7}} \times \sqrt{1 - \frac{1}{4} \cdot \frac{3}{7}} - \sqrt{\frac{3}{7}} \cdot \frac{1}{2} \cdot \sqrt{\frac{3}{7}}$, i.e., $\cos \alpha = 0.5$. Since $0 < \sqrt{\frac{3}{7}} < \frac{\sqrt{2}}{2}$ and $\frac{1}{2} \sqrt{\frac{3}{7}} < \frac{\sqrt{2}}{2}$, then $0 < \arcsin \sqrt{\frac{3}{7}} < \frac{\pi}{4}$ and $0 < \arcsin \frac{1}{2} \sqrt{\frac{3}{7}} < \frac{\pi}{4}$.

But then

$$
0 < \arcsin \sqrt{\frac{3}{7}} + \arcsin \left(\frac{1}{2} \sqrt{\frac{3}{7}}\right) < \frac{\pi}{2}
$$

i.e., $\alpha$ belongs to the first quadrant.

Thus, $\cos \alpha = 0.5$ and $0 < \alpha < \frac{\pi}{2}$, but in this case $\alpha = \frac{\pi}{3}$, and $x_{1} = 0.5 \sqrt{\frac{3}{7}}$ is a root of equation (1).

We check whether $x_{2}$ is a root of equation (1). Let $\beta = \arcsin 2 x_{2} + \arcsin x_{2}$. Then $\arcsin \left(-\sqrt{\frac{3}{7}}\right) + \arcsin \left(-\frac{1}{2} \sqrt{\frac{3}{7}}\right) = \beta$. Since $-1 < -\sqrt{\frac{3}{7}} < 0$ and $-1 < -0.5 \sqrt{\frac{3}{7}} < 0$, then

$-\pi < \arcsin \left(-\sqrt{\frac{3}{7}}\right) + \arcsin \left(-\frac{1}{2} \sqrt{\frac{3}{7}}\right) < 0$, or $-\pi < \beta < 0$.

Therefore, $\beta \neq \frac{\pi}{3}$. Hence, $x_{2}$ is not a root of equation (1).

Answer. $x_{1} = 0.5 \sqrt{\frac{3}{7}} \approx 0.32732683$ (the approximate value is obtained using a calculator).