Consider the case $a > |b|$, so the solutions are unbounded.  The equation is separable, and we get implicit solutions of the form

$$ t + c = \int \frac{dx}{a - b \sin(x)} $$

Now the integrand is periodic with period $2\pi$.  The time to go from $x=x_0$ to $x_0 + 2\pi$ is $$ T = \int_{x_0}^{x_0+2\pi} \dfrac{dx}{a-b\sin(x)}$$
By strict convexity of the function $1/(1-t)$ for $|t| < 1$, 
$$ T < \dfrac{(2 \pi)^2}{\int_{x_0}^{x_0+2\pi} (a - b \sin(x))\; dx} = \frac{2\pi}{a}$$
Thus we can't have $x(t) = a t + o(t)$.