Question

Как решить 3sinx - 2cos^x = -3?

Answer 1

3sinx - 2cos²x = -3
3sinx - 2(1-sin²x) = -3
3sinx - 2 + 2sin²x = -3
2sin²x + 3sinx + 1 = 0
sinx = t
2t² + 3t + 1 = 0
D = 9-8 = 1
√D = 1
x₁ = (-3-1)/4 = -1
x₂ = (-3+1)/4 = -1/2

sinx = -1
sinx = -1/2

x = -π/2 + 2πn
x = -π/6 + 2πn
x = -5π/6 + 2πn
n∈Z