Пожалуйста, помогите с тригонометрией

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