calculus problem?

Question

i need help with this question. please help me with this.

Assume that f(x)=cos^2x.solve f '(x)=0 for x in the interval [0,2pi]

Answer 1

f(x)=cos^2(x)
f'(x)=-2 cos(x) sin(x)

f'(x)=0 whenever either cos or sin is zero;

this occurs when x= 0 [sin(0)=0]
x=pi/2 [cos(pi/2)=0]
x=pi [sin(pi)=0]
x=3 pi/2 [cos[3 (pi/2)=0]
x=2 pi [sin(2 pi)=0]

Answer 2

f'(x) = -2cos(x) * sin(x)
-2cos(x) * sin(x) = 0 => cos(x) = 0 or sin(x) = 0
cos(x) = 0 =>x = pi/2, x = 3/2 * pi
sin(x) = 0 => x = 0, x = pi x = 2pi

so netto : x = 0,x=pi/2,x=pi,x=3/2pi and x = 2pi

Answer 3

f'(x) = -2 cos x sin x
-2 cos x sin x = 0
-sin 2x = 0
x = 0
x = pi/2
x = 2pi

Answer 4

f'(x)=-2sin(2x)=0
sin(2x)=0 in [0,2pi]
2x=0 or 2x=pi or 2x=2pi
x=0 or x=pi/2 or x=2pi