A) 3tg²x + 2tg x - 1 =0 б) ctg²2x - 6ctg 2x + 5 =0 в) 2tg²x + 3tg x - 2 =0 г) 7ctg² x/2 =5

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A) Let y = tg x. The first equation becomes 3y² + 2y - 1 = 0. This is a quadratic equation that can be factored as (3y - 1)(y + 1) = 0. So, y = 1/3 or y = -1.

Substitute y back with tg x:
tg x = 1/3 or tg x = -1

Thus, the solutions are x = arctg(1/3) + kπ or x = arctg(-1) + kπ, where k is an integer.

B) Let y = ctg 2x. The second equation becomes y² - 6y + 5 = 0. This is a quadratic equation that can be factored as (y - 5)(y - 1) = 0. So, y = 5 or y = 1.

Substitute y back with ctg 2x:
ctg 2x = 5 or ctg 2x = 1

Thus, the solutions are x = arctg(1/5) + kπ or x = arctg(1) + kπ, where k is an integer.

C) Let y = tg x. The third equation becomes 2y² + 3y - 2 = 0. This is a quadratic equation that can be factored as (2y - 1)(y + 2) = 0. So, y = 1/2 or y = -2.

Substitute y back with tg x:
tg x = 1/2 or tg x = -2

Thus, the solutions are x = arctg(1/2) + kπ or x = arctg(-2) + kπ, where k is an integer.

D) For the fourth equation, we have 7ctg²(x/2) = 5. This can be rearranged as ctg²(x/2) = 5/7. Taking the square root of both sides, we get ctg(x/2) = ±√(5/7).

The solutions are x/2 = arccot(√(5/7)) + kπ or x/2 = arccot(-√(5/7)) + kπ, where k is an integer. Multiplying both sides by 2 to find the possible values of x.