Solve the following equation:

$$
\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1 .
$$
I. Solution. It is known that $\cos 2 x=2 \cos ^{2} x-1$. We can also express $\cos 3 x$ in terms of $\cos x$:

$$
\begin{gathered}
\cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x= \\
=2 \cos ^{3} x-\cos x-2 \sin ^{2} x \cos x=2 \cos ^{3} x-\cos x- \\
-2\left(1-\cos ^{2} x\right) \cos x=\left(4 \cos ^{2} x-3\right) \cos x
\end{gathered}
$$

With these, (1) becomes

$$
\begin{gathered}
\cos ^{2} x+4 \cos ^{4} x-4 \cos ^{2} x+1+\left(16 \cos ^{4} x-24 \cos ^{2} x+9\right) \cos ^{2} x=1 \\
2 \cos ^{2} x\left(8 \cos ^{4} x-10 \cos ^{2} x+3\right)=0
\end{gathered}
$$

This can be 0 if any of the factors is 0. The disappearance of the second factor gives a quadratic equation in $\cos ^{2} x$. Solving this, we find that (1) is satisfied if

$$
\text { or } \cos ^{2} x=0, \quad \text { or } \quad \cos ^{2} x=\frac{3}{4}, \quad \text { or } \quad \cos ^{2} x=\frac{1}{2}
$$

that is, if

$$
\begin{gathered}
x_{1}=90^{\circ}+n \cdot 180^{\circ}, \quad x_{2}=30^{\circ}+n \cdot 180^{\circ}, \quad x_{3}=150^{\circ}+n \cdot 180^{\circ} \\
x_{4}=45^{\circ}+n \cdot 180^{\circ}, \quad x_{5}=135^{\circ}+n \cdot 180^{\circ} \quad(n \text { is an integer })
\end{gathered}
$$

In the interval $0^{\circ}-360^{\circ}$, the equation has 10 solutions.

The first three and the last two solutions can be combined and written more simply as:

$$
x^{\prime}=30^{\circ}+n \cdot 60^{\circ}, \quad x^{\prime \prime}=45^{\circ}+n \cdot 90^{\circ} \text {. }
$$

Ringler András (Esztergom, I. István g. III. o. t.)

II. Solution. Reducing the equation to 0, the left side can be factored using the repeated application of the identity

$$
2 \cos u \cos v=\cos (u-v)+\cos (u+v)
$$

$$
\begin{aligned}
& \frac{1+\cos 2 x}{2}+\frac{1+\cos 4 x}{2}+\cos ^{2} 3 x-1=\cos 3 x \cos x+\cos ^{2} 3 x= \\
& \quad=\cos 3 x(\cos x+\cos 3 x)=2 \cos 3 x \cos x \cos 2 x=0
\end{aligned}
$$

Thus, the roots can be calculated from the following three equations:

$$
\begin{array}{lll}
\cos 3 x=0, & \cos x=0, & \cos 2 x=0, \text { namely } \\
3 x=90^{\circ}+n \cdot 180^{\circ}, & x=90^{\circ}+n \cdot 180^{\circ}, & 2 x=90^{\circ}+n \cdot 180^{\circ} .
\end{array}
$$

The identity (2) from the first solution shows that $\cos 3 x=0$ also holds alongside $\cos x=0$, so the second equation does not introduce new roots. The solution is again

$$
x=30^{\circ}+n \cdot 60^{\circ}, \quad \text { and } \quad x=45^{\circ}+n \cdot 90^{\circ}
$$

where $n$ is an integer.