Как решить (√ (2) * sinx+1) * (√-5cosx) = 0

(√ (2) * sinx+1) * (√-5cosx) = 0

1) √ (2) * sinx+1=0⇒sinx=-1/√2⇒x=-π/4+2πk U x=-5π/4+2πk, k∈z

2) √-5cosx ≤ 0 ⇒cosx≤0⇒π/2+2πk≤x≤3π/2+2πk, k∈z

Ответ x={π/2+2πk; 3π/2+2πk; -5π/4+2πk, k∈z}