Question

решите неравенства f(x)≤0, если f(x)=sin2x-x

Answer 1

f'(x)=2cos2x-1
2cos2x-1≤0
2cos2x=1
cos2x=1/2
2x=±arccos1/2+2πk
2x=±π/3+2πk
x=±π/6+πk