Problem solving a simple integral equation Could someone help me finding q(t) so that
\begin{equation}
\int_x^{x+2\pi/3}q(t)\,\mathrm{d}t = a\cdot\sin{(x)}+b
\end{equation}
Thanks in advance!

My try: 
\begin{equation}
\int_0^{\frac{2\pi}{3}}q\left(v+x\right)\,\mathrm{d}v = a \sin{x}+b 
\end{equation}
In the other hand:
$$\int_0^{\frac{2\pi}{3}} \frac{3xa}{2\pi}\cos\left(\frac{xv}{2\pi/3}\right) dv=a\sin\left(x\right)$$
$$\int_0^{\frac{2\pi}{3}} \frac{b}{2\pi/3} dv=b$$
Then:
$$\int_0^{\frac{2\pi}{3}}q\left(v+x\right)\,\mathrm{d}v= \int_0^{\frac{2\pi}{3}} \frac{3xa}{2\pi}\cos\left(\frac{xv}{2\pi/3}\right) dv+\int_0^{\frac{2\pi}{3}} \frac{b}{2\pi/3}dv$$
$$\int_0^{\frac{2\pi}{3}} \left( q\left(v+x\right)-\frac{3xa}{2\pi}\cos\left(\frac{xv}{2\pi/3}\right)-\frac{b}{2\pi/3} \right) \,\mathrm{d}v=0$$
