Решение
5sinx+cosx=5
Применяя формулы:
sinx = sin2*(x/2); cosx = cos2*(x/2)
sin²x/2 + cos²x/2 = 1
Получим уравнение:
5* sin2*(x/2) + cos2*(x/2) = 5*(sin²x/2 + cos²x/2)
5*(2sinx/2 * cosx/2) + (cos²x/2 - sin²x/2) = 5*(sin²x/2 + cos²x/2)
10sinx/2 * cosx/2 + cos²x/2 - sin²x/2 - 5sin²x/2 - 5cos²x/2 = 0
- 6sin²x/2 + 10sinx/2 * cosx/2 - 4cos²x/2 = 0 делим на (- 2cos²x/2 ≠ 0)
3tg²x/2 - 5tgx + 2 = 0
tgx = t
3t² - 5t + 2 = 0
D = 25 - 4*3*2 = 1
t₁ = (5 - 1)/6 = 4/6 = 2/3
t₂ = (5 + 1)/6 = 6/6 = 1
tgx = 2/3
x₁ = arctg(2/3) + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πn, n ∈ Z