## Task 11/71

For which real numbers $a$ does the equation

$$
\sin ^{2}(a x)-\cos x+1=0
$$

have exactly one solution?
Due to $\sin ^{2} a x \geq 0$ and $\cos x \leq 1$, it must hold that $\sin ^{2} a x=0$ and $\cos x=1$. Thus, the initial solutions are $a x=k \pi ; x=2 l \pi$ with $k, l$ being integers. This results in the condition for $a$ as the equation $a=\frac{k}{2 l}$.

If $a$ is a real number of the form $\frac{k}{2 l}$, then the equation has infinitely many, not exactly one, solutions. If, on the other hand, $a$ is a real number of the form $\frac{k}{2 l+1}$, then $x=2 l \pi$ results, leading to the contradiction $2 l=2 l+1$; i.e., there is no solution.

The case remains where $a$ is an irrational number. Then the equation is satisfied by the single value $x=0$. Thus, the equation has exactly one solution for all irrational numbers $a$.