What are the points of inflection of #f(x)=cos^2x # on the interval #x in [0,2pi]#?
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To find the points of inflection of ( f(x) = \cos^2(x) ) on the interval ( x \in [0, 2\pi] ), we need to find where the concavity changes. Points of inflection occur where the second derivative changes sign.

Find the first derivative: [ f'(x) = 2\cos(x)\sin(x) ]

Find the second derivative: [ f''(x) = 2(\sin^2(x)  \cos^2(x)) ]

Determine where the second derivative changes sign by finding its zeros: [ 2(\sin^2(x)  \cos^2(x)) = 0 ] [ \sin^2(x)  \cos^2(x) = 0 ] [ \sin^2(x) = \cos^2(x) ] [ \tan^2(x) = 1 ]
Solve for ( x ) in the interval ( x \in [0, 2\pi] ) to find where the concavity changes.
Solutions: ( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} )
So, the points of inflection of ( f(x) = \cos^2(x) ) on the interval ( x \in [0, 2\pi] ) are ( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} ).
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