<s>[INST] Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$. 
Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$ and has no critical points.

For increasing function, $$f'(x)\gt0\\2\cos(2x) - 8(a+1)\cos x + (4a^2 + 8a -14)\gt0$$
Using $\cos(2x)=2\cos^2x-1$, $$4\cos^2x-8(a+1)\cos x+4a^2+8a-16\gt0\\\cos^2x-2(a+1)\cos x+a^2+2a-4\gt0\\(\cos x-(a+1))^2-5\gt0\\(\cos x-a-1+\sqrt5)(\cos x-a-1-\sqrt5)\gt0\\\cos x\lt a+1-\sqrt5 \text{ or }\cos x\gt a+1+\sqrt5$$
Now, I know that $\cos x$ lies between $-1$ and $1$. Not able to use that to solve the inequality.
Also, the same question has been discussed here, but the approach is different. Also, the final answer posted there is not correct.
 [/INST] You are almost done! Each of the two inequalities has to be true for all $x \in \mathbb{R}$.
Let us consider the first situation:
$$\text{cos} \ x<a+1-\sqrt{5} , \ (\forall) \ x \in \mathbb{R}$$
Since the maximum value of $\text{cos} \ x$ is $1$ we get that $$1<a+1-\sqrt{5} \iff \sqrt{5} <a$$
For the second case:
$$\text{cos} \ x>a+1+\sqrt{5} , \ (\forall) \ x\in \mathbb{R}$$
Since the minimum value of $\text{cos} \ x$ is $-1$ we get that $$-1>a+1+\sqrt{5} \iff -2-\sqrt{5} >a$$
Thus, $$a \in (-\infty,-2-\sqrt{5}) \cup (\sqrt{5}, \infty)$$
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