For all $k>0$ suth that $x_k=\frac{\pi}{2}+2k\pi, x_k<1964-189=1775$ we had 2 solutions. It give $2[\frac{\frac{3550}{\pi}+3}{4}]$ positive solutions. By analogy $2[\frac{\frac{4306}{\pi}+3}{4}]$ negative solutions.

You should also prove that $x_k> 1775$ implies no solution (and same for negative values). This is true but must be proved. For example, if we use $192.7032$ instead of $189$, your method misses two negative roots.

You are right. Let us consider $x=a\sin x+b.a>max(1,|b|),a,b -real$. Negative solutions can be found as $-y$, were $y$ is positive solutions $y=a\sin y-b$. Therefore we can consider only positive solutions. Let $x_k=\frac{\pi}{2}(1+4k), y_k=\frac{\pi}{2}(3+4k),a_k=\frac{\pi}{2}(4k)$. Obviosly if $x>a+b$, then $x>asinx+b$. Obviosly between $x_k$ and $y_k$ we had exactly one solution if $x_k<a+b$. Between $y_k$ and $a_{k+1}$ we had one solution if $b>y_k$, else no solution. Between $a_{k+1},x_{k+1}$ we have exactly one solutions when $x_{k+1}>a+b,b<a_{k+1}$. It give calculated number of solutions. But when $x_{k+1}<a+b$ we can have $0$ or $1$ or $2$ solutions and must chek them. This case very specific. We have exactly one solution $x_*$, when $x_*=\sqrt{a^2-1}+b,\cos x_*=\frac{1}{a}$. We have no solution, when $x_*>a\sin x_*+b$. We have 2 solutions, when $x_*<a\sin x_*+b$.