Task 13. Find the minimum value of the function

$$
\psi(x)=\sqrt{15-12 \cos x}+\sqrt{7-4 \sqrt{3} \sin x}
$$

on the interval $[0 ; 0.5 \pi]$.
Solution. Since $|\cos x| \leqslant 1$ and $|\sin x| \leqslant 1$, the function $\psi(x)$ is defined on the entire interval $[0 ; 0.5 \pi]$. To find the properties of the functions $f(x)=\sqrt{15-12 \cos x}$, $\varphi(x)=\sqrt{7-4 \sqrt{3} \sin x}, \quad \psi(x)=f(x)+\varphi(x) \quad$, we create a table of their values using a calculator:

| $x^{0}$ | $f(x)$ | $\varphi(x)$ | $\varphi(x)$ |
| ---: | :---: | :---: | :---: |
| 0 | 1.7320508 | 2.6457513 | 4.3778021 |
| 5 | 1.7451827 | 2.5290645 | 4.2742472 |
| 10 | 1.783921 | 2.407616 | 4.1915837 |
| 15 | 1.8463179 | 2.2818523 | 4.1281702 |
| 20 | 1.9296859 | 2.1518401 | 4.0815260 |
| 25 | 2.030887 | 2.017937 | 4.0487624 |
| 30 | 2.1465542 | 1.8803984 | 4.0269526 |
| 35 | 2.2738018 | 1.7395827 | 4.0133843 |
| 40 | 2.409884 | 1.5958185 | 4.0056669 |
| 45 | 2.5523945 | 1.4494896 | 4.0018841 |
| 50 | 2.6993608 | 1.3010333 | 4.0003941 |
| 55 | 2.8490493 | 1.1509769 | 4.0000262 |
| 60 | 3.000020 | 0.9999899 | 4.0000001 |
| 65 | 3.1509650 | 0.84906731 | 4.0000323 |
| 70 | 3.3008723 | 0.69972737 | 4.0003597 |
| 75 | 3.4487282 | 0.5548598 | 4.0036527 |
| 80 | 3.5939145 | 0.42077535 | 4.0146899 |
| 85 | 3.7355228 | 0.31330512 | 4.0488279 |
| 90 | 3.8729833 | 0.26794924 | 4.1409325 |

Based on the table, we can hypothesize that on the interval $[0 ; 0.5 \pi]$, the function $\psi(x) \geqslant 4$, and $\psi\left(\frac{\pi}{3}\right)=4$.

Thus, let's try to prove that

$$
\sqrt{15-12 \cos x}+\sqrt{7-4 \sqrt{3} \sin x} \geqslant 4
$$

Let $\sqrt{7-4 \sqrt{3} \sin x}=t$.

From this, $\sin x=\frac{7-t^{2}}{4 \sqrt{3}}, \cos x=\sqrt{1-\frac{\left(7-t^{2}\right)^{2}}{48}}$ and the inequality (2) takes the form:

$$
\sqrt{15-\sqrt{3} \cdot \sqrt{48-\left(7-t^{2}\right)^{2}}} \geqslant 4-t
$$

Both sides of the inequality (3) are non-negative, so after squaring both sides, we get an equivalent inequality:

$$
15-\sqrt{3} \cdot \sqrt{48-\left(7-t^{2}\right)^{2}} \geqslant(4-t)^{2}
$$

or

$$
\sqrt{3} \cdot \sqrt{48-\left(7-t^{2}\right)^{2}} \leqslant 8 t-1-t^{2}
$$

From here, $t^{4}-4 t^{3}+6 t^{2}-4 t+1 \geqslant 0$.

Applying the theorem of Bezout, we get $(t-1) \cdot\left(t^{3}-\right.$ $\left.-3 t^{2}+3 t-1\right) \geqslant 0$ or $(t-1)^{4} \geqslant 0$.

Thus, it is proven that the minimum value of the function $\psi(x)$ is 4.