For any real number $b$ let $f(b)$ denote the maximum value of $g(x)$ where $$g(x)=\left|\sin x+\frac{2}{3+\sin x}+b\right|$$
Find the minimum value of $f(b)$
$f(b)$ appears to be lying between $b$ and $b+\frac{3}{2}$. So what can the minimum value of $f(b)$
Let $t=3+\sin x$, then $2\leq t\leq 4$ so $$3\leq t+{2\over t} \leq {9\over 2}$$ and $$g(x)= \underbrace{|t-3+{2\over t}+b|}_{=:h(t)}$$
If $b\geq 0$ then $$h(t) = t+{2\over t}+b-3 \leq {3\over 2}+b = f(b) \implies \min f(b) = {3\over 2}$$
If $b\leq -{3\over 2}$ then $$h(t) = -t-{2\over t}-b+3 \leq -b = f(b) \implies \min f(b) = {3\over 2}$$
If $-{3\over 2}<b<0$ then $$f(b)= \max \{h(2),h(4)\} = \max \{-b,{3\over 2}+b \}$$
$$={{3\over 4} +|b+{3\over 4}| } \implies \boxed{\min f(b) = {3\over 4}}$$ and eqaulity is achieved at $b=-{3\over 4}$
Notice: I use this formula for max at the end $$ \max\{a,b\} = {a+b+|a-b|\over 2}$$