Question

Tg6x−tg2x/1+tg6x⋅tg2x=1/√3

Answer 1

Тоже с якласса?))

Вот, держи. Вместо πn там πk. Ну, кому как удобно.

Answer 2

To solve the trigonometric equation tg(6x) - tg(2x) / (1 + tg(6x) * tg(2x)) = 1/√3, we'll need to manipulate the equation to isolate x and then solve for its values. Here's the step-by-step solution:

Given equation: tg(6x) - tg(2x) / (1 + tg(6x) * tg(2x)) = 1/√3

Step 1: Substitute tan(x) for tg(x): tg(6x) - tg(2x) / (1 + tg(6x) * tg(2x)) = 1/√3 => tan(6x) - tan(2x) / (1 + tan(6x) * tan(2x)) = 1/√3

Step 2: Apply the difference of tangent formula: tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A) * tan(B))

So, the equation becomes: tan(6x - 2x) = 1/√3 => tan(4x) = 1/√3

Step 3: Find the values of x that satisfy the equation: We know that tan(pi/6) = 1/√3 (this is a common trigonometric value).

So, we have: tan(4x) = tan(pi/6)

Since tangent is periodic with a period of π, we can write: 4x = pi/6 + n * pi, where n is an integer.

Solving for x: x = (pi/6 + n * pi) / 4

This gives us an infinite number of solutions for x, where n is any integer.

In summary, the equation has an infinite number of solutions given by x = (pi/6 + n * pi) / 4, where n is an integer. Each value of n will yield a different solution for x that satisfies the equation.