2sinx·sin2x-3sin²x=0;
2sinx·2sinx·cosx-3sin²x=0;
sin²x(4cosx-3)=0;
sin²x=0 или 4cosx-3=0; cosx=3/4
x=πk, k∈Z x=(+-)arccos(3/4)+2πk, k∈Z
-1,5<-V2<-1,4;
5-1,5<5-V2<5-1,4;
3,5<5-V2<3,6.
V знак кв. корня
Решение
Cos²<span>x - sin(3</span>π/2 <span>+ x) = 0
cos</span>²x + cosx = 0
cosx*(cosx + 1) = 0
1) cosx = 0
x₁ = π/2 + πk, k ∈ Z
2) cosx + 1 = 0
cosx = - 1
x₂ = π + 2πn, n ∈ Z
Ответ: x₁ = π/2 + πk, k ∈ Z ; <span>x₂ = π + 2πn, n ∈ Z</span>