Note that $\theta = \cos^{-1}x$ function, by definition, has domain $x\in [-1,1]$ and range in $\theta \in [0,\pi]$, such that
$$\theta =\cos^{-1}x \iff \cos \theta = x$$
we also have
$$\cos (\pi -\theta) = -x$$
and therefore
$$\cos^{-1}(-x)=\pi-\theta \iff \cos (\pi-\theta) = -x$$
that is
$$\cos^{-1}x + \cos^{-1}(-x)=\theta+\pi-\theta= \pi$$
which leads to $\forall x\in [-1,1]$
$$\cos^{-1}x + \cos^{-1}(-x)=\pi$$
In this case we have $\sin \frac{4\pi}{3} = -\frac{\sqrt 3}2$ and $\cos^{-1} \left(\frac{\sqrt 3}2\right)=\frac \pi 6$ then
$$\cos^{-1} \left(\sin \frac{4\pi}{3}\right)=\cos^{-1} \left(-\frac{\sqrt 3}2\right)= \pi-\cos^{-1} \left(\frac{\sqrt 3}2\right)$$
For the first point, by symmetry, from the unit circle, we have that
$$\sin \theta = \cos \left(\frac \pi 2-\theta\right)$$
therefore
$$\sin \frac{4\pi}{3}= \cos \left(-\frac {5\pi} 6\right)=-\cos \left(\frac {\pi} 6\right)=-\frac{\sqrt 3}2$$
or also
$$\sin \theta = \sin (\pi -\theta) \implies \sin \frac{4\pi}{3}= \sin \left(-\frac{\pi}{3}\right)=-\sin \left(\frac{\pi}{3}\right)=-\frac{\sqrt 3}2$$
Best Answer
$f(0)=-4$, $f(\pi/2)=3$ and $$-4\leq f(x)\leq 3,$$
for all $x\in\mathbb R$. Intermediate value theorem gives $f(\mathbb R)=[-4,3]$.