Solve the equation

$$
\cos 10 x+\frac{\tan 5 x}{\cot 5 x}=2
$$
I. solution: By substituting $5 x=y$ and using the identities $\cos 2 y=\cos ^{2} y-\sin ^{2} y=1-2 \sin ^{2} y, \operatorname{ctg} y=1 / \operatorname{tg} y, \operatorname{tg} y=\sin y / \cos y$ and $\cos ^{2} y=1-\sin ^{2} y$, our equation can be transformed as follows:

$$
\begin{aligned}
& 1-2 \sin ^{2} y+\operatorname{tg}^{2} y=2 \\
& 1-2 \sin ^{2} y+\frac{\sin ^{2} y}{1-\sin ^{2} y}=2
\end{aligned}
$$

Let's substitute $\sin ^{2} y$ with $z$ and exclude the possible $z=1$ solution in advance.

$$
1-2 z+\frac{z}{1-z}=2, z^{2}=\frac{1}{2}, z^{2}=\frac{1}{2}, z=\frac{1}{\sqrt{2}}
$$

(The negative root does not apply because $z=\sin ^{2} y \geq 0$). Now we have

$$
\sin y= \pm \frac{1}{\sqrt{2}}= \pm \frac{\sqrt{8}}{2}= \pm 0.8409
$$

From this, with hundredth degree accuracy,

$$
\begin{array}{cc}
y_{1}=57.23^{\circ}+k \cdot 360^{\circ}, & y_{3}=237.23^{\circ}+k \cdot 360^{\circ}, \\
y_{2}=122.77^{\circ}+k \cdot 360^{\circ}, & y_{4}=302.77^{\circ}+k \cdot 360^{\circ},(k=0, \pm 1, \pm 2, \ldots)
\end{array}
$$

Finally, based on $x=y / 5$,

$$
\begin{gathered}
x_{1}=11.45^{\circ}+k \cdot 72^{\circ}, \quad x_{3}=47.45^{\circ}+k \cdot 72^{\circ} \\
x_{2}=24.55^{\circ}+k \cdot 72^{\circ}, \quad x_{4}=60.55^{\circ}+k \cdot 72^{\circ}
\end{gathered}
$$

According to this, our equation has 4 solutions in each $72^{\circ}$ interval determined by the boundaries $0^{\circ}, 72^{\circ}, 144^{\circ}, 216^{\circ}, 288^{\circ}$, and $360^{\circ}$, and 20 solutions between $0^{\circ}$ and $360^{\circ}$.

Katona Éva (Bp. XIV., Ybl M. construction t. I. o. t.)

II. solution: We can directly get an equation for $\cos 10 x$ if we transform the second term in the same way as in the first solution, using the identities $\frac{\operatorname{tg} 5 x}{\operatorname{ctg} 5 x}=\operatorname{tg}^{2} 5 x=\frac{\sin ^{2} 5 x}{\cos ^{2} 5 x}$ and apply the half-angle identities $\sin ^{2} a / 2=(1-\cos \alpha) / 2$ and $\cos ^{2} \alpha / 2=(1+\cos \alpha) / 2$ with $\alpha=10 x$:

$$
\cos 10 x+\frac{1-\cos 10 x}{1+\cos 10 x}=2
$$

From this, assuming $\cos 10 x \neq-1$,

$$
\cos ^{2} 10 x-2 \cos 10 x-1=0, \cos 10 x=1 \pm \sqrt{2}
$$

Only the $\cos 10 x=1-\sqrt{2}=-0.4142$ root applies, as the other is greater than 1. According to the table,

$$
10 x_{1}=114.47^{\circ}+k \cdot 360,10 x_{2}=245.53^{\circ}+k \cdot 360^{\circ},(k=0, \pm 1, \pm 2, \ldots)
$$

and thus, again with hundredth degree accuracy:

$$
x_{1}=11.45^{\circ}+k \cdot 36^{\circ}, \quad x_{2}=24.55^{\circ}+k \cdot 36^{\circ}
$$

Thus, for every integer $k$, there are two roots of our equation between $k \cdot 36^{\circ}$ and $(k+1) \cdot 36^{\circ}$ (and in every other $36^{\circ}$ interval), and between $0^{\circ}$ and $360^{\circ}$, there are ten times as many, or 20 roots.

Losonczy László (Miskolc, Gábor Á. construction t. IV.o. t.)

III. solution: Recalling that every trigonometric function of an angle can be expressed without a square root in terms of the tangent of half the angle, we can transform our equation so that only $\operatorname{tg} 5 x$ appears. Indeed, $\cos 10 x=\left(1 \operatorname{tg}^{2} 5 x\right) /\left(1+\operatorname{tg}^{2} 5 x\right)$ and thus, after multiplying by the non-vanishing $1+\operatorname{tg}^{2} 5 x$ and rearranging,

$$
\operatorname{tg}^{4} 5 x-2 \operatorname{tg}^{2} 5 x-1=0, \quad \operatorname{tg}^{2} 5 x=1 \pm \sqrt{2}
$$

(assuming, of course, that $\operatorname{tg} 5 x$ exists, i.e., $5 x \neq 90^{\circ}+k \cdot 180^{\circ}$). From the positive roots,

$$
\begin{aligned}
\operatorname{tg} 5 x & = \pm \sqrt{1+\sqrt{2}}= \pm 1.5538 \\
5 x & = \pm 57.23^{\circ}+k \cdot 180^{\circ} \quad(k=0, \pm 1, \pm 2, \ldots) \\
x & = \pm 11.45^{\circ}+k \cdot 36^{\circ}
\end{aligned}
$$

From $x_{1}=+11.45^{\circ}$ with $k=0,1,2, \ldots, 9$, and from $x_{2}=-11.45^{\circ}$ with $k=1,2,3, \ldots, 10$, we get angles between $0^{\circ}$ and $360^{\circ}$, and the number of such roots is 20.

Bartha László (Balassagyarmat, Balassi B. g. IV. o. t.)

Remark. The roots (II) with even $k$ give the $x_{1}$ and $x_{3}$ roots of (I), and with odd $k$ the $x_{2}$ and $x_{4}$ roots. The root $x_{2}=-11.45^{\circ}+36^{\circ}$ of (III) appears in (I) as $x_{2}$ with $k=0$.