Here's another argument (assuming it's correct) using the Sturm comparison theorem.

Given $b - \pi < a \le x \le b$, let $$ f(x) = \sin (x - b + \pi). $$ Observe that $f > 0$ and $f'' + f = 0$ on $[a,b)$, $f(a) > 0$, $f(b) = 0$, and $f'(b) = -1$.

Since \begin{align*} (fg'-f'g)' &= (g'' + g)f - (f''+f)g \le 0\\ (fg'-f'g)(b) &= f(b)g'(b) - f'(b)g(b) = g(b) \ge 0, \end{align*} it follows that, on $[a,b]$, $$ fg' - f'g \ge 0. $$ and therefore, on $[a,b)$, \begin{align*} \left(\frac{g}{f}\right)' &= \frac{fg' - gf'}{f^2} \ge 0. \end{align*} Since $$ \frac{g(a)}{f(a)} \ge 0, $$ it follows that, on $[a,b)$, $$ \frac{g}{f} \ge 0. $$

Here's another argument using Sturm-Liouville theory.

Given $b - \pi < a \le x \le b$, let $$ f(x) = \sin (x - b + \pi). $$ Observe that $f > 0$ and $f'' + f = 0$ on $[a,b)$, $f(a) > 0$, $f(b) = 0$, and $f'(b) = -1$.

Since \begin{align*} (fg'-f'g)' &= (g'' + g)f - (f''+f)g \le 0\\ (fg'-f'g)(b) &= f(b)g'(b) - f'(b)g(b) = g(b) \ge 0, \end{align*} it follows that, on $[a,b]$, $$ fg' - f'g \ge 0. $$ and therefore, on $[a,b)$, \begin{align*} \left(\frac{g}{f}\right)' &= \frac{fg' - gf'}{f^2} \ge 0. \end{align*} Since $$ \frac{g(a)}{f(a)} \ge 0, $$ it follows that, on $[a,b)$, $$ \frac{g}{f} \ge 0. $$