Решите уравнение. Найдите все корни, принадлежащие промежутку

cos (4x + Pi/2) =√ 2 cos2x; [-3Pi; -2pi]- sin 4x = √2 cos2x√2 cos2x + 2 sin2x cos2x = 0cos2x ( √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€Zk=-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 cos2xv2 cos2x + 2 sin2x cos2x = 0cos2x ( 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 +pim1) -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 <= -2,25n = -3, x = - 11/4 p2) -3p <= 5р/8 + pik <= -2p-3 <= 5/8 + k <= -2-24 <= 5 + 8k <= -16-29 <= 8k <= -21- 3,625 <= k <= - 2,625k = -3, x = -19/8 р3) 3p <= 7р/8 + pim <= -2p-3 <= 7/8 + m <= -2-24 <= 7 + 8m <= -16-31 <= 8m <= -23- 3,875 <= m <= - 2,875m = -3, x = -17/8 p