## Task 28/75

The equation is to be solved:

$$
\sqrt[3]{1+\lg \tan x}+\sqrt[3]{1-\lg \tan x}=2
$$
If both sides of the given equation are raised to the third power, after simplification and factoring, one obtains the equation

$$
\sqrt[3]{1-\lg ^{2} \tan x}(\sqrt[3]{1+\lg \tan x}+\sqrt[3]{1-\lg \tan x})=2
$$

The expression in parentheses is equal to the left side of the given equation, so it follows that

$$
\sqrt[3]{1-\lg ^{2} \tan x}=1 \rightarrow \lg \tan x=0 \rightarrow \tan x=1 \rightarrow x=\frac{\pi}{4}+k \pi
$$

(where $k$ is any integer).