Ответ(ы) на вопрос:
ГостьРешение
1. Sin2xcosx+cos2xsinx= √3/2
sin(2x + x) = √3/2
sin3x = √3/2
3x = (-1)^x arcsin(√3/2) + πn, n∈ Z
3x = (-1)^n *(π/3) + πn, n∈Z
x = (-1)^n *(π/9) + πn/3, n∈Z
2. Cosx+cos²x= 1/2 - sin²x
cosx = 1/2 - 1
cosx = - 1/2
x = (+ -)arccos(-1/2) + 2πn, n ∈ Z
x = (+ -)*(π - arccos(1/2)) + 2πn, n ∈ Z
x = (+ -)*(π - π/3) + 2πn, n∈Z
x = (+ -)*(2π/3) + 2πn, n∈Z
3. Sin3xcosx-cos3xsinx = √3/2
sin(3x - x) = √3/2
sin2x = √3/2
2x = (-1)^n arcsin(√3/2) + πk, k ∈ Z
2x = (-1)^n * (π/3) + πk, k ∈ Z
x = (-1)^n * (π/6) + πk/2, k ∈ Z
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