# What is #f(x) = int -cos2x dx# if #f(pi/3) = 0 #?

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To find ( f(x) = \int -\cos(2x) , dx ) if ( f(\frac{\pi}{3}) = 0 ), integrate the function ( -\cos(2x) ) with respect to ( x ).

[ \int -\cos(2x) , dx = -\frac{1}{2} \sin(2x) + C ]

Where ( C ) is the constant of integration.

Now, substitute ( \frac{\pi}{3} ) for ( x ) and ( f(\frac{\pi}{3}) = 0 ) into the equation.

[ -\frac{1}{2} \sin\left(2\left(\frac{\pi}{3}\right)\right) + C = 0 ]

[ -\frac{1}{2} \sin\left(\frac{2\pi}{3}\right) + C = 0 ]

[ -\frac{1}{2} \sin\left(\frac{\pi}{3}\right) + C = 0 ]

[ -\frac{1}{2} \left(\frac{\sqrt{3}}{2}\right) + C = 0 ]

[ -\frac{\sqrt{3}}{4} + C = 0 ]

[ C = \frac{\sqrt{3}}{4} ]

So, the function ( f(x) = \int -\cos(2x) , dx ) with ( f(\frac{\pi}{3}) = 0 ) is:

[ f(x) = -\frac{1}{2} \sin(2x) + \frac{\sqrt{3}}{4} ]

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