I'm going to have a look on your try, and then add a claim and an example which might help you.
From
$$-(\cos^2x+\sin^2x)\leqslant 2\cos x\sin x\leqslant \cos^2x+\sin^2x\tag1$$
and
$$\small -(\cos^2 x+\sin^2 x+\sin^2\alpha)\leqslant 2\cos x\sqrt{\sin^2x+\sin^2\alpha}\leqslant \cos^2x+\sin^2x+\sin^2\alpha\tag2$$
you got
$$-2-\sin^2\alpha\leqslant 2f(x)\leqslant 2+\sin^2\alpha$$
and
$$-1-\frac{\sin^2\alpha}{2}\leqslant f(x)\leqslant 1+\frac{\sin^2\alpha}{2}\tag3$$
To see if $(3)$ is the range of $f(x)$, let us see if there is $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$, i.e.
$$2f(x)=2+\sin^2\alpha\tag4$$
holds.
From $(1)(2)$, $x$ satisfying $(4)$ has to satisfy both
$$2\cos x\sin x= \cos^2x+\sin^2x\tag5$$
and
$$2\cos x\sqrt{\sin^2x+\sin^2\alpha}=\cos^2x+\sin^2x+\sin^2\alpha\tag6$$
$(5)$ is equivalent to
$$(\cos x-\sin x)^2=0\iff \cos x=\sin x\iff x=\frac{\pi}{4}+m\pi\ (m\in\mathbb Z)$$
$(6)$ is equivalent to
$$\cos x\geqslant 0\quad\text{and}\quad \cos^2x=\sin^2x+\sin^2\alpha\iff \cos x\geqslant 0\quad\text{and}\quad \cos(2x)=\sin^2\alpha$$
So, we see that
$$(5)\ \ \text{and}\ \ (6)\iff x=\dfrac{\pi}{4}+2m\pi\quad\text{and}\quad \alpha=n\pi\quad (m,n\in\mathbb Z)$$
This means that there is no $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$ holds.
So, $(3)$ is not the range of $f(x)$.
In the following, I'm going to add a claim and an example which might help you.
In the following, I consider only continuous functions.
Claim : If
$f(x)\leqslant F(x)\leqslant g(x)\quad$ ($f(x),g(x)$ are not necessarily constant functions)
$h(x)\leqslant G(x)\leqslant i(x)\quad$ ($h(x),i(x)$ are not necessarily constant functions)
both $f(x)+h(x)$ and $g(x)+i(x)$ are constant functions
there is $\alpha$ such that $F(\alpha)=f(\alpha)$ and $G(\alpha)=h(\alpha)$
there is $\beta$ such that $F(\beta)=g(\beta)$ and $G(\beta)=i(\beta)$
then, $\operatorname{range}(F(x)+G(x))=\left[f(x)+h(x),g(x)+i(x)\right]$.
Proof :
For $c$ such that $f(x)+h(x)\leqslant c\leqslant g(x)+i(x)$, letting $H(x)=F(x)+G(x)-c$, since we have
$$H(\alpha)=F(\alpha)+G(\alpha)-c=f(\alpha)+h(\alpha)-c\leqslant 0$$
$$H(\beta)=F(\beta)+G(\beta)-c=g(\beta)+i(\beta)-c\geqslant 0$$
we can say that, by the intermediate value theorem, there is $\gamma$ such that
$$H(\gamma)=0\qquad \text{and}\qquad \min(\alpha,\beta)\leqslant \gamma\leqslant \max(\alpha,\beta).\quad\blacksquare$$
Exmaple for which the claim above works :
Find the range of $\cos x+\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$.
Let $f(x)=\cos x$ and $g(x)=\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$. Then, $g'(x)=\dfrac{-8 \pi (x-2\pi)(x+\pi)}{3 (x^2 + 2 \pi^2)^2}$ and $\displaystyle\lim_{x\to\pm\infty}g(x)=\frac 13$.
Since $-1\leqslant f(x)\leqslant 1$ and $-1=g(-\pi)\leqslant g(x)\leqslant g(2\pi)=1$, adding these gives
$$-2\leqslant f(x)+g(x)\leqslant 2\tag7$$
Since $f(-\pi)=g(-\pi)=-1$ and $f(2\pi)=g(2\pi)=1$, we can say, by the claim above, that $(7)$ is the range.
Best Answer
Consider $g(x)=\frac{x^2}{\sqrt{x^2+1}}$, an even function that is decreasing for $x\lt 0$ and increasing for $x\gt 0$. We see that
$$g’(x) = \frac{x^3+2x}{(x^2+1)^{\frac 32}} =0 \implies x=0$$ Clearly, this corresponds to a minimum. Therefore, $0\le g(x) \le 1$ where the upper bound is a consequence of the restriction by $\cos^{-1}x$. This means $f(x)=\cos^{-1}(g(x))$ can vary from $\cos^{-1} 0$ to $\cos^{-1} 1$. Indeed, your answer is correct. The range is $$\left[0,\frac{\pi}{2}\right]$$