Question

Найти нули функции у=cosх/2 найти нули функции у=sin 3x

Answer 1

$y=cos\frac{x}{2} \\ \left \{ {{x=0} \atop {y=cos0}} \right. \left \{ {{x=0} \atop {y=1}} \right. \\ \left \{ {{cos\frac{x}{2}=0} \atop {y=0}} \right. \left \{ {{\frac{x}{2}=\frac{\pi}{2}+\pi n,\ n\in Z \atop {y=0}} \right. \left \{ {{x=\pi+2\pi n,\ n\in Z} \atop {y=0}} \right. \\OTBET: \{(0;1);(\pi+2\pi n;0)\},\ n\in Z$
$\\y=sin3x \\ \left \{ {{x=0} \atop {y=sin0}} \right. \left \{ {{x=0} \atop {y=0}} \right. \\ \left \{ {{sin3x=0} \atop {y=0}} \right. \left \{ {{3x=\pi n,\ n\in Z} \atop {y=0}} \right. \left \{ {{x=\frac{\pi n}{3},\ n\in Z} \atop {y=0}} \right. \\OTBET:(\frac{\pi n}{3};0),\ n\in Z$