Sin[latex] sin^{2}x - \frac{5}{2} sin2x +2 =0[/latex]

Решение sin²x - (5/2)*2sinxcosx + 2*sin²x + 2cos²x = 0 3sin²x - 5sinxcosx + 2cos²x = 0   делим на cos²x ≠ 0 3tg²x - 5tgx + 2 = 0 tgx = t 3t² - 5t + 2 = 0 D = 25 - 4*3*2 = 1 t₁ = (5 - 1)/6 t₁ = 2/3 t₂ = (5 + 1)/6 t₂ = 1 1)  tgx = 2/3 x₁ = arctg(2/3) + πk, k ∈ Z 2)  tgx = 1 x₂ = π/4 + πn, n ∈ Z