Cos2x=2cos²x-1
2cos²x-1-2√2cosx-2=0
2cos²x-2√2cosx-3=0
cosx=t
2t²-2√2t-3=0
D=(2√2)²-4*2*(-3)=32, √D=4√2
t1=(2√2+4√2)/4=3√2/2 - не подходит, т.к. больше 1
t2=(2√2-4√2)/4=-√2/2
cosx=-√2/2
x=+-arccos(-√2/2)+2πn
x=+-(π-arccos√2/2)+2πn
x=+-3π/4 +2πn
n=1, x=5π/4 ∈[π;5π/2]