Function involving definite integral

[Saitama]

Homework Statement

Let g be a continuous function on R that satisfies $\displaystyle g(x)+2\int_{0}^{\pi/2} \sin x \cos t g(t)dt=\sin x$, then $g'\left(\frac{\pi}{3}\right)$ is equal to
A)1/2
B)1/√2
C)1/4
D)none of these

Homework Equations

The Attempt at a Solution

Rewriting the given expression,
g(x)=\sin x\left(1-2\int_0^{\pi/2}\cos t g(t)dt \right)
g(x)=k\sin x

where $\displaystyle k=\left(1-2\int_0^{\pi/2}\cos t g(t)dt \right)$.
g'\left(\frac{\pi}{3}\right)=\frac{k}{2}

I am stuck here, how would I evaluate k?

Any help is appreciated. Thanks!

 

[HallsofIvy]

You know that g(t)= k sin(t) so you can do the integral:
k= 1- 2k\int_0^{2\pi} cos(t)sin(t) dt

 

[Saitama]

You know that g(t)= k sin(t) so you can do the integral:
k= 1- 2k\int_0^{2\pi} cos(t)sin(t) dt

Thanks HallsofIvy! Silly me, missed such an obvious step.