Inspired by Shivamani Patil

We note that:

$\begin{aligned} \tan{(x-30^\circ)} & = \dfrac{\sin{(x-30^\circ)}}{\cos{(x-30^\circ)}} = \dfrac{\cos{(120^\circ-x)}} {\sin{(120^\circ-x)}} \\ & =\dfrac{\cos{(x-120^\circ)}} {-\sin{(x-120^\circ)}} = - \cot{(x-120^\circ)} \\ \Rightarrow \dfrac{\tan{(x+120^\circ)} }{\tan{(x-30^\circ)}} & = -\tan{(x+120^\circ)} \tan{(x-120^\circ)} \\ & = - \frac{\tan{x}-\sqrt{3}}{1+\sqrt{3} \tan{x}} \dot{} \frac{\tan{x}+\sqrt{3}}{1-\sqrt{3} \tan{x}} \\ & = - \frac{\tan^2{x}-3}{1-3 \tan^2{x}} = \frac{11}{2} \\ \Rightarrow - 2\tan^2{x} - 6 & = 11 - 33 \tan^2{x} \\ \Rightarrow \tan^2{x} & = \frac{5}{31} \end{aligned}$

We know that:

$\begin{aligned} \cos{2x} & = \frac{1 - \tan^2{x}}{1+\tan^2{x}} = \frac{1-\frac{5}{31}}{1+\frac{5}{31}} = \frac{26}{36} = \frac{13}{18} \\ \Rightarrow \cos{4x} & = 2\cos^2{2x} - 1 = 2\left( \frac{13}{18} \right)^2 - 1 = \frac{7}{162}\end{aligned}$

$\Rightarrow a + b = 7 + 162 = \boxed{169}$

Using the identity $\tan(x + y) = \dfrac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)},$ and the facts that $\tan(120^{\circ}) = -\sqrt{3}$ and $\tan(30^{\circ}) = \dfrac{1}{\sqrt{3}}$ , the given equation can be written as

$\dfrac{\dfrac{\tan(x) - \sqrt{3}}{1 + \sqrt{3}\tan(x)}}{\dfrac{\tan(x) - \dfrac{1}{\sqrt{3}}}{1 + \dfrac{\tan(x)}{\sqrt{3}}}} = \dfrac{11}{2}$

$\Longrightarrow \dfrac{(\tan(x) - \sqrt{3})(\tan(x) + \sqrt{3})}{(1 + \sqrt{3}\tan(x))(\sqrt{3}\tan(x) - 1)} = \dfrac{\tan^{2}(x) - 3}{3\tan^{2}(x) - 1} = \dfrac{11}{2}$

$\Longrightarrow 2\tan^{2}(x) - 6 = 33\tan^{2}(x) - 11 \Longrightarrow \tan^{2}(x) = \dfrac{5}{31}.$

Now $\cos(4x) = 2\cos^{2}(2x) - 1,$ and

$\cos(2x) = 2\cos^2(x) - 1 = \dfrac{2}{\sec^{2}(x)} - 1 = \dfrac{2}{1 + \tan^{2}(x)} - 1 =$

$\dfrac{2}{1 + \dfrac{5}{31}} - 1 = \dfrac{31}{18} - 1 = \dfrac{13}{18}.$

Thus $\cos(4x) = 2*\left(\dfrac{13}{18}\right)^{2} - 1 = \dfrac{169}{162} - 1 = \dfrac{7}{162},$

and so $a + b = 7 + 162 = \boxed{169}.$

If $\frac { \tan { (x+120) } }{ \tan { (x-30) } } =\frac { m }{ n }$

Then by ratio and proportions we have

$\frac { m+n }{ m-n } =\frac { \tan { (x+120) } +\tan { (x-30) } }{ \tan { (x+120) } -\tan { (x-30) } }$

$=\frac { \frac { \sin { (x+120) } }{ \cos { (x+120) } } +\frac { \sin { (x-30) } }{ \cos { (x-30) } } }{ \frac { \sin { (x+120) } }{ \cos { (x+120) } } -\frac { \sin { (x-30) } }{ \cos { (x-30) } } }$

$=\frac { \frac { \sin { (x+120)\cos { (x-30) } +\sin { (x-30)\cos { (x+120) } } } }{ \cos { (x-30) } \cos { (x+120) } } }{ \frac { \sin { (x+120)\cos { (x-30) } -\sin { (x-30)\cos { (x+120) } } } }{ \cos { (x-30) } \cos { (x+120) } } }$

$=\frac { \sin { (x+120)\cos { (x-30) } +\sin { (x-30)\cos { (x+120) } } } }{ \sin { (x+120)\cos { (x-30) } -\sin { (x-30)\cos { (x+120) } } } }$

$=\frac { \sin { (2x+90) } }{ \sin { (150) } } =2\sin { (2x+90) }$

$=2(\sin { 2x } \cos { 90 } +\sin { 90 } \cos { 2x } )=2\cos { 2x }$

$\therefore \cos { 2x } =\frac { m+n }{ 2(m-n) }$

$\therefore \cos { 4x } =2{ (\cos { 2x } ) }^{ 2 }-1=2\times { \left( \frac { 13 }{ 18 } \right) }^{ 2 }-1=\frac { 7 }{ 162 }$

[tan(x + 120)]/[tan(x - 30)] = {[sin(x + 120)]/[cos(x + 120)]} {[cos(x - 30)]/[sin(x - 30)]} = [4 cos^2(x) - 1]/[4 cos(2(x) - 3] = 11/2,so cos^2(x) = 31/36. cos(4x) = 8 cos^4(x) - 8 cos^2(x) + 1 = 8 (31/36)^2 - 8*(31/36) + 1 = (7688 - 8928 + 1296)/1296 = 56/1296 = 7/162, and 7 + 162 = 169