Решить уравнение и сделать отбор корней: (√2sinx+1)*√(-5cosx)=0 [ -5π; -7π/2 ]

ОДЗ
cosx≤0⇒x∈[π/2+2πn;3π/2+2πn,n∈z]
cosx=0⇒x=π/2+πn,n∈z
-5π≤π/2+πn≤-7π/2
-10≤1+2n≤-7
-11≤2n≤-8
-5,5≤n≤-4
n=-5⇒x=π/2-5π=-9π/2
n=-4⇒x=π/2-4π=-7π/2
√2sinx+1=0
sinx=-1/√2
x=5π/4+2πk,k∈z U x=7π/4+2πm,m∈z
-5π≤5π/4+2πk≤-7π/2
-20≤5+8k≤-14
-25≤8k≤-19
-25/8≤k≤-19/8
k=-3⇒x=5π/4-6π=-19π/4
-5π≤7π/4+2πm≤-7π/2
-20≤7+8k≤-14
-27≤8k≤-21
-27/8≤k≤-21/8
k=-3⇒x=7π/4-6π=-17π/4 не удовл условию,т.к. сos(-17π/4)>0