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