Answer
$-\frac{7}{2}$ $\lt$ $a$ $\lt$ $-\frac{5}{2}$
Work Step by Step
$f(x)$ = $(a^{2}+a-6)\cos 2x+(a-2)x+\cos 1$
Determine the first derivative:
$f'(x)$ = $-(a^{2}+a-6)(2\sin 2x)+(a-2)$
The derivative exists for all $x$, so the only possible critical points will occur where
$f'(x)$ = $0$
$-2(a^{2}+a-6)\sin 2x+(a-2)$ = $0$
$2(a-2)(a+3)\sin 2x$ $- (a-2)$=$0$
$(a-2)(2(a+3)\sin 2x -1)=0$
$a = 2$ or $2(a+3)\sin 2x = 1$
$a = 2$ or $\sin 2x = \frac{1}{2(a+3)}$
Since the range of $\sin 2x$ is $[-1,1]$, the second equation has no solution whenever either
$\frac{1}{2(a+3)}$ $\lt$ $-1$ or $\frac{1}{2(a+3)}$ $\gt$ $1$
so
$-\frac{7}{2}$ $\lt$ $a$ $\lt$ $-\frac{5}{2}$
