Range of an inverse trigonometric function Find the range of $f(x)=\arccos\sqrt {x^2+3x+1}+\arccos\sqrt {x^2+3x}$
My attempt is:I first found domain,
$x^2+3x\geq0$
$x\leq-3$ or $x\geq0$...........(1)
$x^2+3x+1\geq0$
$x\leq\frac{-3-\sqrt5}{2}$ or $x\geq \frac{-3+\sqrt5}{2}$...........(2)
From (1) and (2),
domain is $x\leq-3$ or $x\geq0$
but could not solve further..Any help will be greatly appreciated.

Notice, 
for the defined function $\cos^{-1}\sqrt{x^2+3x+1}$$$\implies -1\leq \sqrt{x^2+3x+1}\leq 1$$ $$\implies x^2+3x+1\geq 1$$ $$\implies x^2+3x\geq 0$$ $$\implies x(x+3)\geq 0$$ The above inequality holds for all $x$ such that $$x\in (-\infty, -3]\cup [0, \infty)\tag 1$$
Again, for the defined function $\cos^{-1}\sqrt{x^2+3x}$$$\implies -1\leq \sqrt{x^2+3x}\leq 1$$ $$\implies x^2+3x\geq 1$$ $$\implies x^2+3x-1 \geq 0$$ 
Solving the quadratic equation $x^2+3x-1=0$ for $x$, we have  $$\left(x-\frac{-3+\sqrt{13}}{2}\right)\left(x-\frac{-3-\sqrt{13}}{2}\right)\geq 0$$ The above inequality holds for all $x$ such that $$x\in\left(-\infty, \frac{-3-\sqrt{13}}{2}\right]\cup \left[\frac{-3+\sqrt{13}}{2}, \infty\right)\tag 2$$ Form (1) & (2), we get domain of function $f(x)$ $$\color{blue}{x\in (-\infty, -3]\cup [0, \infty)}$$
Now, setting extreme point $x=-3$ in $f(x)$, we get 
$$f(-3)=\cos^{-1}\sqrt{(-3)^2+3(-3)+1}+\cos^{-1}\sqrt{(-3)^2+3(-3)}$$ $$=\cos^{-1}(1)+\cos^{-1}(0)=\frac{\pi}{2}$$ Setting extreme point $x=0$ & $x=0$ in $f(x)$, we get $$f(0)=\cos^{-1}\sqrt{(0)^2+3(0)+1}+\cos^{-1}\sqrt{(0)^2+3(0)}$$ $$=\cos^{-1}(1)+\cos^{-1}(0)=\frac{\pi}{2}$$ In both cases we get equal values hence the range of $f(x)$ is $\color{blue}{\frac{\pi}{2}}$
