For the inner function
$$\sin^{-1} s +\frac 23 (\frac{\pi}{2}-\sin^{-1} x)$$
$$\frac 13 (\pi +\sin^{-1}x)$$ $$\frac{-\sin^{-1} x}{3}$$
Since it is inside a log function which is inside a square root $$-\frac{\sin^{-1} x}{3} \ge 1$$
$$\sin^{-1}x \le -3$$ which is looks wrong, and is in fact wrong, because the answer is $[0,\sqrt{\log (\frac{\pi}{2})}]$
Where am I going wrong?
On
Since $$\arcsin{x}+\arccos{x}=\frac{\pi}{2}$$ and $$0\leq\arccos{x}\leq\pi,$$ we obtain: $$\frac{\pi}{6}=\frac{\pi}{2}-\frac{\pi}{3}\leq\arcsin{x}+\frac{2}{3}\arccos{x}=\frac{\pi}{2}-\frac{1}{3}\arccos{x}\leq\frac{\pi}{2}-\frac{1}{3}\cdot0=\frac{\pi}{2}.$$ But by the domain should be $$\arcsin{x}+\frac{2}{3}\arccos{x}\geq1$$ and since $\ln$ increases, we obtain the answer: $$\left[0,\sqrt{\ln\frac{\pi}{2}}\right].$$
You’re wrong in saying that $$\pi+\sin^{-1} x =-\sin^{-1}x $$ We have, $$-\frac{\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2} \\ \frac{\pi}{6} \le \frac 13 (\pi +\sin^{-1} x ) \le \frac{\pi}{2} $$Since $\log(x)$ is increasing, $${\log\frac{\pi}{6}} \le f^2(x) \le \log\frac{\pi}{2} $$ Now we need to take the square root. For this to be defined we must have the argument inside $\log$ to be $\ge 1$, and so the range comes out to be $$0\le f(x)\le \sqrt{\log\frac{\pi}{2}} $$