What is $f (x) = \int - cos 2 x d x$ $f(x) = int -cos2x dx$ if $f (\frac{π}{3}) = 0$ $f(pi/3) = 0$ ?

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Answer 1 · 2016-10-06T18:51:22

$f (x) = - \frac{1}{2} sin 2 x + \frac{\sqrt{3}}{4} .$ $f(x)=-1/2sin2x+sqrt3/4.$

$f (x) = \int - cos 2 x d x = - \frac{sin 2 x}{2} + C$ $f(x)=int-cos2xdx=-(sin2x)/2+C$

To determine $C$ $C$ , we make use of the cond. that, $f (\frac{π}{3}) = 0$ $f(pi/3)=0$

$f (\frac{π}{2}) = 0 \Rightarrow - \frac{1}{2} sin (2 (\frac{π}{3})) + C = 0$ $f(pi/2)=0 rArr -1/2sin(2(pi/3))+C=0$

$:. C=1/2sin(pi-pi/3)=1/2sin(pi/3)=1/2*sqrt3/2=sqrt3/4$

$:. f(x)=-1/2sin2x+sqrt3/4.$

What is f(x) = int -cos2x dxf(x)=∫−cos2xdxf(x) = int -cos2x dx if f(pi/3) = 0 f(π3)=0f(pi/3) = 0 ?