Verified answer

2sin²x - 3sinx*cosx - 5cos²x = 0 / cos²x

2tg²x - 3tgx - 5 = 0

Пусть tgx = t. Тогда:

2t² - 3t - 5 = 0

D = b² - 4ac = 9 - 4 * 2 * (-5) = 49

t₁ = (- b - √D)/2a = (3 - 7)/4 = -1

t₂ = (- b + √D)/2a = (3 + 7)/4 = 2 1/2

Вернемся к замене:

tgx = t

tgx = - 1

x = 3π/4 + πk; k ∈ Z

tgx = 2 1/2

x = arctg (2 1/2) + πk; k ∈ Z

Ответ: x = arctg (2 1/2) + πk; k ∈ Z; 3π/4 + πk; k ∈ Z