Solve the following equation:

$$
\cos x + \cos 3x = \frac{1}{2} \tan 4x
$$
Based on known identities, the left side can be factored, the right side can be expressed as a fraction, and then further transformed (1) as follows:

$$
2 \cos x \cos 2 x=\frac{\sin 4 x}{2 \cos 4 x}=\frac{\sin 2 x \cos 2 x}{\cos 4 x}=\frac{2 \sin x \cos x \cos 2 x}{\cos 4 x}
$$

and by factoring the difference of the two sides:

$$
2 \cos x \cos 2 x\left(1-\frac{\sin x}{\cos 4 x}\right)=0
$$

This holds if

$$
\begin{aligned}
& \cos x=0, \quad \text { i.e., } \quad x_{1}=(2 k-1) \cdot 90^{\circ}, \text { and if } \\
& \cos 2 x=0, \quad \text { i.e., } \quad x_{2}=(2 k-1) \cdot 45^{\circ}
\end{aligned}
$$

(since the denominator is non-zero in both cases); and also if

$$
\frac{\sin x}{\cos 4 x}=1, \quad \sin x=\cos 4 x=\sin \left(90^{\circ}-4 x\right)
$$

This holds if

$$
\begin{aligned}
& x-\left(90^{\circ}-4 x\right)=k \cdot 360^{\circ}, \text { and if } \\
& x+\left(90^{\circ}-4 x\right)=(2 k+1) 180^{\circ}
\end{aligned}
$$

i.e.,

$$
\begin{gathered}
x_{3}=(4 k+1) \cdot 18^{\circ}=18^{\circ}+k \cdot 72^{\circ} \\
x_{4}=30^{\circ}-(2 k+1) \cdot 60^{\circ}=-30^{\circ}+k \cdot 120^{\circ}=(4 k-1) \cdot 30^{\circ}
\end{gathered}
$$

(in each solution, $\cos 4 x \neq 0$).

According to these, in the interval $0^{\circ}, 360^{\circ}$, (1) is satisfied by the following values:

$$
x=18^{\circ}, 45^{\circ}, 90^{\circ}, 135^{\circ}, 162^{\circ}, 210^{\circ}, 225^{\circ}, 234^{\circ}, 270^{\circ}, 306^{\circ}, 315^{\circ}, 330^{\circ}
$$

(and if $x$ satisfies the equation, then $x+k \cdot 360^{\circ}$ also satisfies it).

By marking the intersection point of the moving arm of the corresponding angle with the unit circle, we obtain the vertices of a regular pentagon and a regular triangle, one of whose vertices is on the positive $y$-axis, as well as the vertices of a square whose sides are parallel to the axes, and finally the other endpoint of the diameter on the $y$-axis (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9effb010bf4cbcbebe3g-1.jpg?height=580&width=547&top_left_y=1737&top_left_x=759)

József Závoti (Székesfehérvár, Ybl M. Gimn., IV. o. t.)

Note. For $x=18^{\circ}$, each term of (1) is positive, and on both sides of the equality, expressions of the symmetry axis of the regular pentagon are given (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9effb010bf4cbcbebe3g-2.jpg?height=526&width=444&top_left_y=141&top_left_x=838)