# How do you find the critical numbers of #y = sin^2 x#?

We know that #cos2x=cos^2x-sin^2x=>cos2x=1-2*sin^2x=> sin^2x=1/2[1-cos2x]#

where n is an integer.

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To find the critical numbers of ( y = \sin^2(x) ), you first need to find the derivative of the function with respect to ( x ), and then solve for values of ( x ) where the derivative is equal to zero or undefined.

First, find the derivative of ( \sin^2(x) ) using the chain rule. The derivative of ( \sin^2(x) ) with respect to ( x ) is ( 2\sin(x)\cos(x) ).

Next, set the derivative equal to zero and solve for ( x ).

( 2\sin(x)\cos(x) = 0 )

This equation is satisfied when either ( \sin(x) = 0 ) or ( \cos(x) = 0 ).

The solutions for ( \sin(x) = 0 ) occur at ( x = k\pi ), where ( k ) is an integer.

The solutions for ( \cos(x) = 0 ) occur at ( x = \frac{\pi}{2} + k\pi ) and ( x = \frac{3\pi}{2} + k\pi ), where ( k ) is an integer.

These values of ( x ) are the critical numbers of the function ( y = \sin^2(x) ).

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