finding a function from given function here is a function for:
$f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$
what is the $f(x)$?
I calculate $f(x)$ as follows:
$$\begin{align}
x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\
3f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6} \\ 
f(\frac{\pi}{3}) &=(1/3)\sin\frac{5\pi}{6}
\end{align}$$
$f(x)=(1/3)\sin\frac{5x}{2}$
 A: Assuming $f$ is defined for all $x\in\mathbb{R}$. First, note that for any $x$,
$$
f(x) = \sin\!\left(x+\frac{\pi}{2}\right)-2f\!\left(\frac{\pi}{3}\right) = \cos x -2f\!\left(\frac{\pi}{3}\right)
$$
so it only remains to compute $f\!\left(\frac{\pi}{3}\right)$. From the expression above
$$
f\!\left(\frac{\pi}{3}\right) = \cos \frac{\pi}{3} -2f\!\left(\frac{\pi}{3}\right) = \frac{1}{2} -2f\!\left(\frac{\pi}{3}\right)
$$
and therefore rearranging the terms gives $f\!\left(\frac{\pi}{3}\right) = \frac{1}{6}$. Putting it all together,
$$
\forall x\in \mathbb{R}, \quad f(x)=\cos x - \frac{1}{3}\;.
$$
(It then only remains to check this expression satisfies the original functional equation, to be certain. It does; but even a quick sanity check for $x=0$, $x=\frac{\pi}{2}$ and $x=\pi$ will be enough to build confidence.)
A: $f(x)=f((x+\frac{1}{2}\pi)-\frac{1}{2}\pi)=\sin(x+\frac{1}{2}\pi)-2f(\frac{1}{3}\pi)=\cos(x)-2f(\frac{1}{3}\pi)$
To find $f(\frac{1}{3}\pi)$ we substitute $x=\frac{1}{3}\pi$:
$f(\frac{1}{3}\pi)=\cos(\frac{1}{3}\pi)-2f(\frac{1}{3}\pi)$
Then $f(\frac{1}{3}\pi)=\frac{1}{3}\cos(\frac{1}{3}\pi)=\frac{1}{6}$ so we end up with:
$$f(x)=\cos(x)-\frac{1}{3}$$
A: Shift the argument by $\pi/2$:
$$f(x)=\sin(x+\frac\pi2)-2f(\frac{\pi}3).$$
Then plug $\pi/3$:
$$f(\frac{\pi}3)=\sin(\frac{5\pi}6)-2f(\frac{\pi}3).$$
Solve for $f(\pi/3)$:
$$f(x)=\sin(x+\frac\pi2)-\frac23\sin(\frac{5\pi}6).$$
