ОДЗ
-5cosx≥0⇒cosx≤0⇒x∈[π/2+2πn;3π/2+2πn,n∈z]
√2sinx+1=0
sinx=-1/√2
x=-π/4+2πn,n∈z
-5π≤-π/4+2πn≥-7π/2
-20≤-1+8n≤-14
-19≤8n≤-13
-19/8≤n≤-13/8
n=-2⇒x=-π/4-4π=-47π/4
x=5π/4+2πn не удов ОДЗ
сos=0⇒x=π/2+πn,n∈z
-5π≤π/2+πn≤-7π/2
-10≤1+2n≤-7
-11≤2n≤-8
-11/2≤n≤-4
n=-5⇒x=π/2-5π=-9π/2
n=-4⇒x=π/2-4π=-7π/2