Question

6sin^2(pi-x)-1,5cos(pi/2-2x)-cos^2x=1

Answer 1

6sin²(π - x) - 1,5cos(π/2 - 2x) - cos²x = 1

6sin²x - 1,5sin2x - cos²x = 1

6sin²x - 1,5•2•sinx•cosx - cos²x = sin²x + cos²x

5sin²x - 3•sinx•cosx - 2cos²x = 0

Разделим обе части cos²x ≠ 0

5tg²x - 3tgx - 2 = 0

Замена tgx = a, a ∈ R

5a² - 3a - 2 = 0

D = (-3)² - 4•5•(-2) = 9 + 40 = 49

a₁ = (3 - 7)/10 = - 4/10 = - 2/5  ⇒ tgx = - 2/5 ⇔ x = - arctg(2/5) + πn, n ∈ Z

a₂ = (3 + 7)/10 = 10/10 = 1  ⇒ tgx = 1 ⇔ x = π/4 + πk, k ∈ Z

ОТВЕТ: - arctg(2/5) + πn ; π/4 + πk , n , k ∈ Z