Values of $a$ for which $ f(x)=8 a x-a \sin6x -7x - \sin 5x $ increases Please help me in this question:
Find all the values of the parameter $a$ for which the function
$f(x)= 8 a x-a \sin6 x -7 x - \sin 5 x $ increases and has no critical points for all real $x$.
I start with $f'(x)= (8 a -6 a \cos6 x -7-5\cos5 x) >0.$ What to do 
further?
 A: You want to know for which $a\in{\mathbb R}$ we have
$$f_a'(x)=a\bigl(8-6\cos(6x)\bigr)-\bigl(7+5\cos(5x)\bigr)\geq0$$
for all $x\in{\mathbb R}$. 
Since $f_a'(0)=2a-12$ we have the necessary condition $a\geq6$. In order to show that this is also sufficient we write $f_a'$ in the following way:
$$f_a'(x)=2(a-6)+12a\sin^2(3x)+10\sin^2{5x\over2}\ .$$
Note that when $a\geq6$ then $f_a'(x)\geq0$ for all $x$, and even if $a=6$ one has $f_a'(x)=0$ only in isolated points. This guarantees that $f_a$ is strictly  increasing when $a\geq6$.
A: $\textbf{Hint}$ 
$8a -6a\cos6x -7-5\cos5x>0 \implies a(8-6\cos {6x}) > 7 + 5\cos{5x} \implies a > \large{\frac{7 + 5\cos{5x}}{8 - 6\cos{6x}}} $
A: Hint: Solve for $a$ to get an inequality of the form $a > g(x)$.
This inequality must be true for all values of $x$. For what values of $a$ is this true for $g(x) = \sin x$? For $g(x) = x^2$? How can you find the valid $a$s for arbitrary $g$?
A: Treat $a$ as a variable. Let $ a = y $.
Draw 
$ y= f(x) = \dfrac{7 x + \sin 5 x}{ 8 x - \sin 5 x} $ 
or implicit function 
$ 8 x y - y \sin (6 x) - 7 x - \sin ( 5 x ) = 0 $  on x and y  axes. 
it has a shape somewhat like the sinc function  $ y= sin(x)/ x $ curve. It is the needed demarcation curve.
In first and third quadrants it is positive and in second and forth negative.
