Решение
1) sin(π - x) - cos(3π/2 - x) = √2
sinx + cosx = √2
преобразуем левую часть уравнения:
sinx + cosx = √2(cosx/√2 + sinx/√2) =
√2[sin(π/4) * cosx + cos(π/4) * sinx] = √2sin(π/4 + x)
Получаем уравнение:
√2sin(π/4 + x) = √2
делим на √2, получаем:
sin(π/4 + x) = 1
π/4 + x = π/2 + 2πn, n ∈ Z
x = π/4 + 2πn, n ∈ Z
2) sinα = - 24/25 π < α < 3π/2
cosα = - √(1 - sin²α) = - √(1 - (24/25)²) = - √(1 - (576/625)) = - √(49/625) = - 7/25
3) sin²x + 8sinx - 9 = 0
sinx = t, IsinxI ≤ 1
t² + 8t - 9 = 0
t₁ = - 9 не удовлетворяет условию IsinxI ≤ 1
t₂ = 1
sinx = 1
x = π/2 + 2πk, k ∈ Z
4) 2cos²x - 9cosx - 5 = 0
cosx = t, IcosxI ≤ 1
2t² - 9t - 5 = 0
D = 81 + 4*2*5 = 121
t₁ = (9 - 11)/4 = - 1/2
t₂ = (9 + 11)/4 = 5 не удовлетворяет условию IcosxI ≤ 1
cosx = - 1/2
x = (+ -)arccos(- 1/2) + 2πn, n ∈ Z
x = (+ -) * (π - π/3)+ 2πn, n ∈ Z
x = (+ -) * (2π/3)+ 2πn, n ∈ Z