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 = $\frac{3 + 23}{20}$ = $\frac{26}{20} = \frac{13}{10}$ - не уд. условию.
t₂ = $\frac{3 - 23}{20} = \frac{-20}{20} = -1$
Обратная замена:
$cos2x = -1$
$2x = \pi + 2 \pi n$ , n ∈ Z.
$x = \frac{ \pi }{2} + \pi n$ , n ∈ Z.