Решение
4sinx+5cosx=4
Применяя формулы:
sinx = sin2*(x/2); cosx = cos2*(x/2)
sin²x/2 + cos²x/2 = 1
Получим уравнение:
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