To find the values of a for $f$ has no critical number Determine the values of $a$ for which $$f(x)=(a^2+a-6)\cos{2x}+(a-2)x+\cos{1}$$has no critical number
My attempt:
A critical number of $f$ is a number $c$ in the domain of $f$ such that $f'(c)=0$ or $f'(c)$ does not exist
$$f'(x)=-2(a^2+a-6)\sin{2x}+(a-2)$$
So,
$$f'(0)=0$$
gives $a=2$
But how do I find the other values?

We need
$$f'(x)=-2(a^2+a-6)\sin{2x}+(a-2)=0$$
Then
$$f'(x)=-2(a+3)(a-2)\sin{2x}+(a-2)$$
Because we want that $f$ has no critical points, we need that $f'(x)$ is never $0$. But $-2(a+3)(a-2)\sin{2x}$ can be $0$ (since $\sin{2x}=0$ an infinite number times in the domain).
So really, we need that $(a-2)$ is never $0$, so that even when $-2(a+3)(a-2)\sin{2x}=0$, no critical point is achieved. And $(a-2)$ is never $0$ as long as $x \not = 2$.
Furthermore, we need to exclude values of $a$ such that $2(a+3)(a-2)\sin{2x} = (a-2)$. Divide by $(a-2)$ to get $$2(a+3)\sin{2x} = 1$$
Solve for $a$ to get $$a = \frac 12 (\csc(2x) - 6)$$ Which means that if $a$ can be expressed in this form for some $x$, then $f$ will have critical points. So we need that $a \not = \frac 12 (\csc(2x) - 6)$ for all $x$.
For ease of notation, let $g(x) = \frac 12 (\csc(2x) - 6)$. The range of $g$ is every real number except for those between $- \frac 72$ and $- \frac 52$, exclusive. So if $a$ is in the excluded range, it will never equal $g(x)$. In other words, if $a \in (- \frac 72, - \frac 52)$, then we have satisfied the criterion that $2(a+3)(a-2)\sin{2x} \not = (a-2)$.
Combining these results gives us that $f$ has no critical points when $a \in (- \frac 72, - \frac 52)$ and $a \not = 2$, which is the same as saying
$$a \in (- \frac 72, - \frac 52)$$
