At what points in the interval `[0,""""2pi]` , does the function `sin2x` attain its maximum value?

Explanation

Let `f(x) = sin 2x rArr f'(x) = 2 cos 2x`
For maxima/minima, f'(x) = 0
` rArr 2 cos 2x = 0 rArr cos 2x = 0 `
`rArr 2x = pi/2,(3pi)/2,(5pi)/2, (7pi)/2`
` rArr x= pi/4, (3pi)/4, (5pi)/4, (7pi)/4`
So we find the values of f(x) at `x = pi/4, (3pi)/4, (5pi)/4,(7pi)/4`
and at the end points of the interval `[0, 2pi]`.
at ` x= 0, f(0) = sin (2 xx 0) =0`
at` x = 2pi, f(2 pi) = sin (2 xx 2pi) = 0`
at ` x = pi/4, f(pi/4) = sin (2 xxpi/4) = sin. pi/2 = 1`
at ` x=(3pi)/4, f((3pi)/4)= sin (2 xx (3pi)/4) = sin. (3pi)/2`
` = sin (pi+pi/2) =- sin. pi/2 =- 1`
at` x= (5pi)/4, f ((5pi)/4)= sin (2 xx(5pi)/4) = sin. (5pi)/2 `
`=sin(2pi+pi/2)=sin.pi/2=1`
at` x=(7pi)/4,f((7pi)/4)=sin(2xx(7pi)/4)=sin.(7pi)/2`
`=sin(2pi+(3pi)/2) = sin. (3pi)/2 =-1`
`:. ` f(x) has absolute maximum value 1 in the interval `[0, 2pi]" which is at "x=pi/4 and x = (5pi)/4`.