Doubly Radical Angle

$\begin{aligned} \cos x & = \frac 12 \sqrt{2+\sqrt{2+\sqrt 3}} \\ \cos^2 x & = \frac {2+\sqrt{2+\sqrt 3}}4 \\ 4 \cos^2 x & = 2+\sqrt{2+\sqrt 3} \\ 4 \cos^2 x - 2 & = \sqrt{2+\sqrt 3} \\ 2(2 \cos^2 x - 1) & = \sqrt{2+\sqrt 3} \\ 2 \cos 2x & = \sqrt{2+\sqrt 3} \\ 4 \cos^2 2x & = 2+\sqrt 3 \\ 4 \cos^2 2x - 2 & = \sqrt 3 \\ 2 \cos 4x & = \sqrt 3 \\ \cos 4x & = \frac {\sqrt 3}2 \\ \implies 4x & = \frac \pi 6 \\ x & = \frac \pi {24} \end{aligned}$

$\implies A = \boxed{24}$

$\large \cos \dfrac \pi A =\dfrac12 \sqrt{2 + \sqrt{2 + \sqrt3}} \implies$

${\bf 4cos^2(\frac{\pi}{A}) = }$ ${\bf 2 + \sqrt{2 + \sqrt{3}} \implies }$

${\bf 4cos^2(\frac{\pi}{A}) - 2 = \sqrt{2 + \sqrt{3}} \implies }$

${\bf 2(2cos^2(\frac{\pi}{A}) - 1) = \sqrt{2 + \sqrt{3}} \implies }$

${\bf 2cos(\frac{2\pi}{A}) = \sqrt{2 + \sqrt{3}} \implies }$

${\bf 4cos^2(\frac{2\pi}{A}) = 2 + \sqrt{3} \implies}$

${\bf 4cos^2(\frac{2\pi}{A}) - 2 = \sqrt{3} \implies}$

${\bf 2(2cos^2(\frac{2\pi}{A}) - 1) = \sqrt{3} \implies }$

${\bf 2cos(\frac{4\pi}{A}) = \sqrt{3} \implies }$

${\bf cos(\frac{4\pi}{A}) = \frac{\sqrt{3}}{2} \implies }$

${\bf \frac{4\pi}{A} = \frac{pi}{6} \implies }$

${\bf A = 24 }$

Finding $x$ , the straight way, would be a long journey finding inverse cos (arccos). Let's find the value of $x$ , through inverse cos formula first.

$\cos(x) = \dfrac{1}{2}\sqrt[]{2+\sqrt[]{2+\sqrt[]{3}}} \\ x = \cos^{-1}\Bigr\{ \dfrac{1}{2}\sqrt[]{2+\sqrt[]{2+\sqrt[]{3}}} \Bigl\} \\ \text{Let } \dfrac{1}{2}\sqrt[]{2+\sqrt[]{2+\sqrt[]{3}}} = y \\ x = \cos^{-1}(y) \\ \text{Using inverse cos formula} \\ \cos^{-1}(A) = \tan^{-1}\Bigl[\dfrac{\sqrt[]{1-A^2}}{A}\Bigr] \\ \implies x = \cos^{-1}(y) = \tan^{-1}\Bigl[\dfrac{\sqrt[]{1-y^2}}{y}\Bigr] \\ \text{Then we find value of arctan using Taylor series} \\ y = \dfrac{1}{2}\sqrt[]{2+\sqrt[]{2+\sqrt[]{3}}} \approx 0.9914 \\ \text{Now let's attempt to find value of } \cos^{-1}(y) \\ \tan^{-1}\Bigl[\dfrac{\sqrt[]{1-y^2}}{y}\Bigr] = \tan^{-1}\Bigl[\dfrac{\sqrt[]{1-(0.9914)^2}}{0.9914}\Bigr] \\ = \tan^{-1}\Bigl[\dfrac{\sqrt[]{0.01712604}}{0.9914}\Bigr] = \tan^{-1}\Bigl[\dfrac{0.1309}{0.9914}\Bigr] = \tan^{-1}(0.132) \\ \text{Therefore, } \\ \tan^{-1}(0.132) = 0.132 - \dfrac{(0.132)^3}{3} + \dfrac{(0.132)^5}{5} - \dfrac{(0.132)^7}{7} + \dfrac{(0.132)^9}{9} \\ \implies \tan^{-1}(0.132) \approx 0.1312 \text{ rad} \\ \color{#3D99F6}{\boxed{\therefore x = \cos^{-1}(y) = 0.1312 \text{ rad }}} \\ \implies x \approx 7.5 ^{\circ} \\ \color{magenta}{\boxed{\therefore x = \dfrac{\pi}{24}} \implies A = 24}$

You can also do this the other way, coming in reverse starting with $\cos(\dfrac{15}{2})$ or $\cos(\dfrac{\pi}{24})$