2(cos 4х)^− 6(cos 2х)^+ 1 = 0

x = ± π/6 + πm/2, m∈Z

Пошаговое объяснение:

Объяснение:

2cos²4x - 6cos²2x + 1 = 0

2(2cos²2x - 1)² - 6cos²2x + 1 = 0

2(4cos⁴2x - 4cos²2x + 1) - 6cos²2x + 1 = 0

8cos⁴2x - 8cos²2x + 2 - 6cos²2x + 1 = 0

8cos⁴2x - 14cos²2x + 3 = 0

cos²2x = t

8t² - 14t + 3 = 0

D/4 = 7² - 24 = 49 - 24 = 25

t_1=\dfrac{7+5}{8}=\dfrac{3}{2}t

1

=

8

7+5

=

2

3

t_2=\dfrac{7-5}{8}=\dfrac{1}{4}t

2

=

8

7−5

=

4

1

cos²2x = 3/2 - нет корней

или

cos²2x = 1/4

cos2x = ± 1/2

cos2x = 1/2 cos2x = - 1/2

2x = ± π/3 + 2πn, n∈Z 2x = ± 2π/3 + 2πk, k∈Z

Корни можно объединить:

2x = ± π/3 + πm

x = ± π/6 + πm/2, m∈Z