Trickonometry II

$\tan 3\theta=\tan\left(\frac{\pi}{3}-x\right)\left(\tan x\right)\tan\left(\frac{\pi}{3}+x\right)$ $=\left(\red{\tan x}\right)\cdot\dfrac{3-\red{\tan^{2}x}}{1-3\red{\tan^{2}x}}=\dfrac{11}{2}$ $\Rightarrow \dfrac{\red{x}\left(3-\red{x^{2}}\right)}{1-3\red{x^{2}}}=\dfrac{11}{2}$ $\Rightarrow 6\red{x}-2\red{x}^{3}=11-33\red{x}^{2}$ $\Rightarrow 2x^{3}-33x^{2}-6x+11=0$ $x=\dfrac{1}{2}\space is\space a\space trivial\space solution$ $\therefore \left(x-\frac{1}{2}\right)\space is\space a\space factor$ $\frac{\left(2x^{3}-33x^{2}-6x+11\right)}{\left(x-\frac{1}{2}\right)}=2\left(x^{2}-16x-11\right)=0$ $\Rightarrow x^2-16x-11=0$ $From\space Quadratic\space Formula$ $x=\dfrac{16{^+_-}\sqrt{16^2-4\times(-11)}}{2}$ $=\dfrac{16{^+_-}10\sqrt{3}}{2}=8{^+_-}\sqrt{75}$ $\Rightarrow x\in\{8+\sqrt{75},8{-}\sqrt{75},0.5\}$ $\because \dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{2}$ $\Rightarrow 1\leq\tan\theta$ $\Rightarrow 1\leq x$ $\Rightarrow x=8+\sqrt{75}$ $\Rightarrow a=8,b=75$ $\Rightarrow a+b=75+8=\boxed{83}$