Finding conditions for a with given condition for critical points 
$f(x)=\sin2x-8(a+1)\sin x+(4a^2+8a-14$)$x$. $x$ increases for all $x \in \mathbb{R}$ and has no critical points. Find values of $a$.

My try:
$f'(x)=4(\cos^2x-2(a+1)\cos x+a^2+2a-4)=0$
and $f''(x)=0$. And on solving i will get range of $a$. That's my answer. But answer doesn't match with original values of answer. Am i missing anthing?
 A: Hint : I am afraid that your derivative is wrong. It is supposed to be  $$2 \cos (2 x)-8 (a+1) \cos (x)$$. Nox, since you are told that $f$ has no critical points, the derivative should not cancel for any $x$ and this will give conditions for $a$. For sure, you must expand $cos(2x)$ as a function of $cos(x)$ to get a quadratic equation.  
I am sure you can take from here.
A: f'($x$)=4[$\mathrm cos^2x$-2($a$+1)$cosx$+$a^2$+2$a$-4]
Let $\phi$($t$) = $t^2$-2($a$+1)$t$+$a^2$+2$a$-4. -1$\le$$t$$\le$1.
It is sufficient to find values of t when $\phi$($t$) = 0 has no roots in [-1,1].
Discriminant D = 20.
USING CONDITIONS FOR QUADRATIC EQUATIONS 
Case I:-     $\phi$(-1) $\gt$ 0.$a$+1 $\le$ -1.D=20.$a$ $\in$ (-$\infty$,-2-$\sqrt5$)$\cup$(-2+$\sqrt5$,+$\infty$) and $a$$\lt$-2.$\implies$ $a$ $\in$(-$\infty$,-2-$\sqrt5$).
Case II:-$\phi$(1) $\gt$ 0.$a$+1 $\gt$ 1.D=20.
$\implies$ $a$ $\in$ (-$\infty$,-$\sqrt5$) $\cup$ ($\sqrt5$,+$\infty$),$a$ $\gt$0 $\implies$ $a$ $\in$ ($\sqrt5$,+$\infty$).Hence final values of $a$ are $a$ $\in$ (-$\infty$,-2-$\sqrt5$) $\cup$ ($\sqrt5$,+$\infty$)
