Find the intervals in which the function `sin 3x, x in [0, pi/2]`, is (a) increasing (b) decreasing

Explanation

Finding `f'(x) `
`f'(x) =d(sin3x)/dx `
`f'(x)=cos3x times 3`
`f'(x)=3.cos3x `
Putting `f'(x)=0`
`3cos3x=0`
`cos3x=0`
We know that `costheta=0`
When `theta=pi/2 &3pi/2`
So , for `cos3x=0`
`3x=pi/2` & `3x=3pi/2`
`x=pi/6 & x=pi/2`
Since `x=pii/6 in [0,pi/2]` & `x=pi/2 in [0,pi/2]`
Both values of x are valid.
So, point `x=pi/6` divides the interval into two disjoint intervals `[0,pi/6)` and `(pi/6,pi/2]`
Checking sign of `f'(x)`
`f'(x)=3.cos3x `
Case1 : `3cos3x>0`
`cos3x>0`

`=>3x in [0,pi/2]`
`=>x in[0,pi/6]`Here in this interval the given function is increasing function
case2:`3.cos3x<0`
`cos3x<0`
`3x in[pi/2,(3pi)/2]`
`x in [pi/6,pi/2]`Here in this interval the given function is decreasing