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.