## Task 20/71

Determine all solutions of the trigonometric equation

$$
\sin ^{70} x+\frac{1}{13} \cos ^{13} x=1
$$
For $|\sin x|=1$, $\sin^{70} x=1$ and $\cos x=0$, so $\frac{1}{13} \cos^{13} x=0$. This initially yields the solutions

$$
x_{0 k}=\frac{\pi}{2}+k \pi \quad \text{with } k= \pm 1 ; \pm 2 ; \pm 3 ; \ldots
$$

Assume there is at least one additional solution $x_{1 k}$. Then

$$
x_{1 k} \neq \frac{\pi}{2}+k \pi, \quad|\sin x_{1 k}|<1, \quad|\cos x_{1 k}| \leq 1
$$

and thus

$$
\sin^{70} x_{1 k}<\sin^{2} x_{1 k} ; \quad \frac{1}{13} \cos^{13} x_{1 k}<\cos^{2} x_{1 k}
$$

Therefore,

$$
\sin^{70} x_{1 k}+\frac{1}{13} \cos^{13} x_{1 k}<\sin^{2} x_{1 k}+\cos^{2} x_{1 k}=1
$$

which contradicts the assumption derived from the problem statement. The solution $x_{k}=\frac{\pi}{2}+k \pi$ is thus the only one.