Решите уравнение. Найдите все корни, принадлежащие промежутку

cos (4x + Pi/2) =√ 2 cos2x; [-3Pi; -2pi]
- sin 4x = √2 cos2x
√2 cos2x + 2 sin2x cos2x = 0
cos2x ( √2 + 2 sin2x) = 0

[ cos2x = 0 ,
[ sin2x = -√2/2,

1) cos2x = 0, 2x=Pi/2+Pin, n€Z,

x=Pi/4+Pin/2, n€Z,

x=-9Pi/4, x=-11Pi/4, € [-3Pi; -2pi]
....
2)sin2x = -√2/2,

2x=(-1)^(k +1) *Pi/4 +Pik, k€Z,

x=(-1)^(k +1) *Pi/8 +Pik/2, k€Z,

x1=-Pi/8 +Pik, k€Z,
x2= 5Pi/8 +Pik, k€Z,

Отбираем х€ [-3Pi; -2pi]

-3Pi≤-Pi/8 +Pik≤-2pi, k€Z,
-3+1/8≤k≤-2+1/8, k€Z,

k=-2, x=-17Pi/8;
....

-3Pi≤5Pi/8 +Pik≤-2pi, k€Z
-3-5/8 ≤k≤-2 -5/8; k€Z

k=-3, x=-19Pi/8
....
Ответ: a)Pi/4+Pin/2, n€Z,;(-1)^(k +1) *Pi/8 +Pik/2, k€Z,

b)-19Pi/8;-17Pi/8;-11Pi/4;-9Pi/4,

cos (4x + pi/2) = v2 cos2x
- sin 4x = v2 cos2x
v2 cos2x + 2 sin2x cos2x = 0
cos2x ( v2 + 2 sin2x) = 0
[ cos2x = 0 …………… х = р/4 + рn
[ sin2x = -v2/2
2х = 5р/4 + 2pik ……… x = 5р/8 + pik
2х = 7р/4 + 2pim ………х = 7р/8 +pim
1) -3p <= р/4 + рn <= -2p
-3 <= 1/4 + n <= -2
-12 <= 1 + 4n <= -8
-13 <= 4n <= -9
-13/4 <= n <= -9/4
-3,25 <= n <= -2,25
n = -3, x = - 11/4 p
2) -3p <= 5р/8 + pik <= -2p
-3 <= 5/8 + k <= -2
-24 <= 5 + 8k <= -16
-29 <= 8k <= -21
- 3,625 <= k <= - 2,625
k = -3, x = -19/8 р
3) 3p <= 7р/8 + pim <= -2p
-3 <= 7/8 + m <= -2
-24 <= 7 + 8m <= -16
-31 <= 8m <= -23
- 3,875 <= m <= - 2,875
m = -3, x = -17/8 p