Решите уравнение √2 cos²x=sin(п/2+x) найдите корни принадлежащие промежутку -7п/2; -2п

√2cos²x-cosx=0
1)cosx=0
x=π/2 +πn,n∈Z
-7π/2≤π/2+πn≤-2π
-4≤n≤-3/2
n=-4,-3,-2.
x1=-7π/2; x2=-5π/2; x3=-3π/2.
2)cosx=√2/2
x=плюс минус π/4+2πn,n∈Z
a)-7π/2≤π/4+2πn≤-2π
-15/8≤n≤-9/8
n=-1
x4=-7π/4
б)-7π/2≤-π/4+2πn≤-2π
-13/8≤n≤-7/8
n=-1
x5=-9π/4