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