Помогите решить тригонометрические уравнения 1) 18tg^2x+3tgx-10=0

2) 5cosx-2sinx=0

3) 4cos^2x+3cosx=0

1) 18tg²x + 3tgx - 10 = 0

Пусть t = tgx.

18t² + 3t - 10 = 0

D = 9 + 4•10•18 = 729 = 27²

t1 = (-3 + 27) / 36 = 24/36 = 2/3

t2 = (-3 - 27) / 36 = - 30/36 = - 5/6

Обратная замена:

tgx = 2/3

x = arctg (2/3) + πn, n ∈ Z

tgx = - 5/6

x = arctg (-5/6) + πn, n ∈ Z.

2) 5cosx - 2sinx = 0

-2tgx + 5 = 0

-2tgx = - 5

tgx = 2/5

x = arctg (2/5) + πn, n ∈ Z.

3) 4cos²x + 3cosx = 0

cosx (4cosx + 3) = 0

cosx = 0

x = π/2 + πn, n ∈ Z

4cosx + 3 = 0

4cosx = - 3

cosx = - 3/4

x = ±arccos (-3/4) + 2πn.