√2cos²x+cosx=0
cosx(√2cosx-+1)=0
cosx=0⇒x=π/2+πn,n∈z
-5π/2≤π/2+πn≤-π
-5≤1+2n≤-2
-6≤2n≤-3
-3≤n≤-1,5
n=-3⇒x=π/2-3π=-5π/2
n=-2⇒x=π/2-2π=-3π/2
cosx=-1/√2⇒x=-2π/3+2πk,k∈z U x=2π/3+2πm,m∈z
-5π/2≤-2π/3+2πk≤-π
-15≤-4+12k≤-6
-11≤12k≤-2
-11/12≤k≤-1/6
нет решения
-5π/2≤2π/3+2πm≤-π
-15≤4+12m≤-6
-19≤12m≤-10
-19/12≤m≤-5/6
m=-1⇒x=2π/3-2π=-4π/3