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

$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$

which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$ and $b$)

$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (-b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}(-b)^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$

Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.

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

$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$

which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$ and $b$)

$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}b^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$

Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.

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

$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$

which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$)

$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (-b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}(-b)^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$

Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.

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

$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$

which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$ and $b$)

$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (-b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}(-b)^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$

Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.

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 $x(t) = a t + o(t)$.

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

$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$

which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$)

$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (-b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}(-b)^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$

Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.