Question

Решите неравенство f'(x) < 0, если f(x)=sin(2x)cos(x) + cos(2x)sin(x)

Answer 1

$\displaystyle\bf\\f(x)=Sin2x Cosx+Cos2x Sinx=Sin(2x+x)=Sin3x\\\\f'(x)=(Sin3x)'=Cos3x\cdot(3x)'=3Cos3x\\\\3Cos3x < 0\\\\Cos3x < 0\\\\\frac{\pi }{2} +2\pi n < 3x < \frac{3\pi }{2} +2\pi n,n\in Z\\\\\\\frac{\pi }{6} +\frac{2\pi n}{3} < x < \frac{\pi }{2} +\frac{2\pi n}{3} ,n\in Z\\\\\\Otvet:x\in\Big(\frac{\pi }{6} +\frac{2\pi n}{3} \ ; \ \frac{\pi }{2}+\frac{2\pi n}{3}\Big) \ ,n\in Z$