2cos^(2)x-3sinx=0
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решение на фото.............

2Cos²x - 3Sinx = 0

Cos²x + Sin² = 1 => Cos²x = 1 - Sin²x

2(1 - Sin²x) - 3Sinx = 0

2 - 2Sin²x - 3Sinx = 0

-2Sin²x - 3Sinx + 2 = 0

2Sin²x + 3Sinx - 2 = 0

Sinx = t ∈ [-1;1]

2t² + 3t - 2 = 0

D = 9 - 4 * 2 * (-2) = 25

t₁ = (-3 + √25) / 4 = 1/2

t₂ = (-3 - √25) / 4 = -2 ∉ [-1;1]

Sinx = 1/2

x = π/6 + 2πn, n∈Z

x = 5π/6 + 2πn, n∈Z