Question

Find the interval in which $f\left(x\right)=\sin 3x-\cos 3x,0<x<\pi$ is strictly increasing or strictly decreasing.

Answer 1

$f\left(x\right)=\sin 3x-\cos 3x$

$f\left(x\right)=3\cos 3x+3\sin 3x$

$f\left(x\right)=0$ gives $\tan 3x=-1$

$x=\pi /4,x=7\pi /12$

In $\left(0,\pi /4\right),f\left(x\right)>00<x<\pi /4$

In $(\frac{\pi }{4},>\frac{\pi }{12})),f\left(x\right)<0$

In $(\frac{7\pi }{12},\pi ),f\left(x\right)>0$

$f\left(x\right)$ is increasing in $(0,\frac{\pi }{4})\cup (\frac{7\pi }{12},\pi )$

And decreasing in $(\frac{\pi }{4},\frac{7\pi }{12})$