Tg(x+пи/3) - sqrt3 0

Предыдущий
вопрос Следующий
вопрос

To solve the inequality Tg(x+π/3) - √3 < 0, we need to isolate the tangent function.

First, let's move the square root term to the other side of the inequality:

Tg(x+π/3) < √3

Next, we need to find the possible values of x for which the tangent of x+π/3 is less than √3. We can start by finding the period of the tangent function, which is π. This means that the tangent function repeats every π units.

Since we are looking for the values of x in terms of x+π/3, we can rewrite the inequality as:
Tg(x) < √3, for x in the interval [0, π].

The tangent function is less than √3 in the intervals (0, π/3) and (2π/3, π).

Therefore, the solution to the inequality Tg(x+π/3) - √3 < 0 is:
x ∈ (0, π/3) U (2π/3, π)