Any Dirichlet series with periodic coefficients is analytically continuable to the whole plane (maybe with a pole at $1$).
It's a finite linear combination of series like $$\sum_{m=1}^\infty\frac1{(km+r)^s}$$ where $1\le r\le k$ which equals $$k^{-s}\sum_{m=1}^\infty\frac1{(m+r/k)^s}.$$ This latter sum is an example of a Hurwitz zeta function well-known to have an analytic continuation. http://en.wikipedia.org/wiki/Hurwitz_zeta_function
Added Looking carefully at your question, I note that despite your title, your series does not actually have periodic coefficients unless $x$ is rational. In general your $k$-th coefficient is $$a_k=\sin^2 2\pi kx=\frac{2-\exp(4\pi i x)-\exp(-4\pi i x)}4.$$ Thus your series can be expressed in terms of the Riemann zeta function and functions of the form $$f_y(s)=\sum_{n=1}^\infty\frac{\exp(2\pi iky)}{n^s}.$$ In effect this sort of function is dual to the Hurwitz zeta function, and it has an analytic continuation to the complex plane with a pole at $1$ proved in the same manner as the Hurwitz zeta function. In the Wikipedia page it has a brief appearance as essentially $\beta(x;s)$. One can express $f_y(1-s)$ in terms of the Hurwitz zeta function.
Any Dirichlet series with periodic coefficients is analytically continuable to the whole plane (maybe with a pole at $1$).
It's a finite linear combination of series like $$\sum_{m=1}^\infty\frac1{(km+r)^s}$$ where $1\le r\le k$ which equals $$k^{-s}\sum_{m=1}^\infty\frac1{(m+r/k)^s}.$$ This latter sum is an example of a Hurwitz zeta function well-known to have an analytic continuation. http://en.wikipedia.org/wiki/Hurwitz_zeta_function