ТРИГОНОМЕТРИЯ. СРОЧНО! Решите уравнение: 3cos2x - 5 cos4x + 8 = 0

3cos2x - 5 cos4x + 8 = 03cos2x - (5 - 10sin²2x) + 8 = 0 3cos2x - 5 + 10sin²2x + 8 = 0 3cos2x + 10 - 10cos²2x + 3 = 0 -10cos²2x + 3cos2x + 13 = 0 10cos²2x - 3cos2x - 13 = 0 Пусть t = cos2x, t ∈ [-1; 1] 10t² - 3t - 13 = 0 D = 9 + 13*10*4 = 529 = 23² t₁ = [latex] \frac{3 + 23}{20} [/latex] = [latex] \frac{26}{20} = \frac{13}{10} [/latex] - не уд. условию. t₂ = [latex] \frac{3 - 23}{20} = \frac{-20}{20} = -1[/latex] Обратная замена: [latex]cos2x = -1 [/latex] [latex]2x = \pi + 2 \pi n[/latex], n ∈ Z. [latex]x = \frac{ \pi }{2} + \pi n[/latex], n ∈ Z.