Ответ:
cos2x + 3sinx - 2 = 0
cos²x - sin²x + 3sinx - 2 =0
1-sin²x - sin²x + 3sinx - 2 = 0
-2sin²x + 3sinx - 1 = 0 |*(-1)
2sin²x - 3sinx + 1 =0
Обозначим: sinx=t,тогда
2t²-3t+1 =0
D= 9-8 =1
t₁= 1, t₂ = 1/2
(1) sinx= 1
x₁= π/2+2πn, n ∈ z
(2) sinx= 1/2
x₂= (-1)^k arcsin1/2 + πk
x₂= (-1)^k π/6 + πk, k∈z
б) x₁= π/2+2πn, n ∈ z
n=1, x= π/2+2π= 5π/2 ∈ [π;5π/2]
x₂= (-1)^k π/6 + πk, k∈z
n= 2, x= (-1)² π/6 +2π = π/6+2π = 13π/6 ∈ [π;5π/2]
Объяснение: