Question

sin(2x+3п)sin(3x+3п/2)=0

Answer 1

Ответ:

$\left \{ {{ x=\frac{\pi k}{2}} \atop {x = \frac{\pi}{6} +\frac{\pi k}{3}}} \right. \, \, \, , k \in Z$

Объяснение:

$sin(2x+3\pi)=-sin2x\\sin(3x+\frac{3\pi}{2})=-cos3x \\\\sin(2x+3\pi)sin(3x+\frac{3\pi}{2})=0\\\\sin2xcos3x=0$

1:

$sin2x=0\\x=\frac{\pi k}{2} \, \, \, , k \in Z$

2:

$cos3x=0\\3x=\frac{\pi}{2} +\pi k\, \, \, , k \in Z\\\\x = \frac{\pi}{6} +\frac{\pi k}{3}\, \, \, , k \in Z$