Question

Срочно помогите решить
(5cos4x - sin4x)2 + (cos4x + 5sin4x)2

Answer 1

Ответ:

4

Объяснение:

4* sin2*(x/2) + 5*cos2*(x/2) = 4*(sin²x/2 + cos²x/2)

4*(2sinx/2 * cosx/2) + 5*(cos²x/2 - sin²x/2) = 4*(sin²x/2 + cos²x/2)

8*sinx/2 * cosx/2 + 5*cos²x/2 - 5*sin²x/2 -  4sin²x/2 - 4cos²x/2 = 0

- 9sin²x/2  + 8sinx/2 * cosx/2  + cos²x/2 = 0 делим на (- cos²x/2 ≠ 0)

9tg²x/2 - 8tgx - 1 = 0

tgx = t

9t² - 8t - 1 = 0

D = 64 + 4*9*1 = 100

t₁ = (8 - 10)/18 = - 2/18 = - 1/9

t₂ = (8 + 10)/18 = 18/18 = 1

tgx = - 1/9

x₁ = arctg(- 1/9) + πk, k ∈ Z

x₁ = - arctg(1/9) + πk, k ∈ Z

tgx = 1

x₂ = π/4 + πn, n ∈ Z