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Ответ

Ответ дан
nKrynka
Решение
2cos²x - 9sinx + 3 = 0
2*(1 - sin²x) - 9sinx + 3 = 0
2sin²x + 9sinx - 5 = 0
Пусть sinx = t, ItI ≤ 1, тогда
2t² + 9t - 5 = 0
D = 81 + 4*2*5 = 121
t₁ = (- 9 - 11)/4
t₁ = - 5 не удовлетворяет условию  ItI ≤ 1
t₂ = (- 9 + 11)/4
t₂ = 1/2
sinx = 1/2
x = (-1)^n*arcsin(1/2) + πn, n∈Z
x = (-1)^n*(π/6) + πn, n∈Z


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