Solve the following equation:

$$
x^{\lg \tan x}+x^{\lg \cot x}=2
$$
I. Solution: Using the relationship $\operatorname{cotg} x=(\operatorname{tg} x)^{-1}$, our equation can be written as:

$$
x^{\lg \operatorname{tg} x}+\frac{1}{x^{\lg \operatorname{tg} x}}=2
$$

Let $x^{\lg \operatorname{tg} x}=y$, then

$$
y+\frac{1}{y}=2
$$

which simplifies to

$$
y^{2}-2 y+1=(y-1)^{2}=0
$$

from which

$$
y_{1}=y_{2}=1
$$

thus

$$
x^{\lg \operatorname{tg} x}=1
$$

Taking the logarithm of both sides

$$
(\lg \operatorname{tg} x) \lg x=\lg 1=0
$$

Therefore, either

$$
\lg x=0, \quad \text { from which } \quad x_{1}=1
$$

or

$$
\lg \operatorname{tg} x=0, \quad \text { from which } \quad \operatorname{tg} x_{2}=1
$$

and thus

$$
x_{2}=\frac{\pi}{4} \pm k \pi, \quad \text { where } \quad k=0,1,2, \ldots
$$

Lackner Györgyi (Bp., V., Fonó-, szövő- és hurkolóip. techn. III. o. t.)

# II. Solution: 

$$
x^{\lg \operatorname{tg} x} \cdot x^{\lg \operatorname{cotg} x}=x^{\lg \operatorname{tg} x+\lg \operatorname{cotg} x}=x^{\lg (\operatorname{tg} x \operatorname{cotg} x)}=x^{\lg 1}=x^{0}=1
$$

Our equation can be written as:

$$
\frac{x^{\lg \operatorname{tg} x}+x^{\lg \operatorname{cotg} x}}{2}=1
$$

From (1) and (2), it is clear that the two terms on the left side of our equation are geometric and arithmetic means, which implies that the two terms are equal and, in this case, each is equal to 1, i.e., $x^{\lg } \operatorname{tg} x=1$ etc., as in the first solution.

Bártfai Pál (Bp., I., Petőfi g. III. o. t.)