For what values of $c$ does the following equation have exactly four solutions:

$$
1+\sin ^{2} c x=\cos x
$$
$$
1+\sin ^{2} c x=\cos x
$$

We see that the number $x=0$ is a solution to the equation for any $c$. The task is thus to determine for which values of $c$ the equation has no non-zero solution.

Since $\sin ^{2} c x$ is certainly non-negative and the maximum value of $\cos x$ is 1, (1) can only be satisfied if

$$
\sin ^{2} c x=0 \quad \text { and } \quad \cos x=1
$$

hold simultaneously, that is,

$$
\begin{array}{rccc}
c x=k \pi, & \text { where } & k & \text { is an integer } \\
x=2 m \pi, & \text { where } & m & \text { is an integer. }
\end{array}
$$

If the equation has a non-zero solution, then according to (3), $m \neq 0$, and substituting (3) into (2) we get

$$
c=\frac{k}{2 m}
$$

which means $c$ is a rational number.

However, if $c$ is a rational number, then the equation also has a non-zero solution: if $c=\frac{p}{q}$, then $x=2 q \pi$ satisfies the equation ($q \neq 0$, since $c$ appears in the denominator):

$$
1+\sin ^{2}\left(\frac{p}{q} 2 q \pi\right)=1+\sin ^{2}(2 p \pi)=1=\cos (2 q \pi)
$$

Thus, we have found that the (1) equation has a non-zero solution if and only if $c$ is a rational number. This implies that the equation has exactly one solution if and only if $c$ is irrational.

Maria Horvath (Hódmezővásárhely, Bethlen G. Gymnasium, 4th grade)