<s>[INST] Solve $\tan x =\sec 42^\circ +\sqrt{3}$ For the trigonometric equation,
$$\tan x =\sec 42^\circ+\sqrt{3}$$
Find the angle $x$, where $0<x<180^\circ$.
I tried to solve for an unknown angle $x$ in a geometry problem with a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached my calculator, entering the right-hand-side and arctan-ing it for $x$.
To my surprise, the angle $x$ comes out at exactly 72 degrees. I did not expect such a neat relationship. Then, I thought I should have solved the equation analytically for the whole-degree angle without the calculator. I spent a good amount of time already and  was not able to derive it yet.
Either the equation is not as innocent as it looks, or a straightforward method just eludes me.
 [/INST] Golden triangles,  I use these mnemonic:
$\;\displaystyle \cos 36° = \frac{\sqrt{2+\frac{1}{ϕ}}}{2} = \frac{ϕ}{2}\;,\;
\sin 36° = \frac{\sqrt{2-\frac{1}{ϕ}}}{2}$
$\;\displaystyle \cos 18° = \frac{\sqrt{2+ϕ}}{2} \qquad\;\;,\;
\sin 18° = \frac{\sqrt{2-ϕ}}{2} = \frac{1}{2ϕ}$

$\displaystyle \cos (60°-18°) = 
\frac{1}{2} × \frac{\sqrt{2+ϕ}}{2} \;+\;
\frac{\sqrt{3}}{2} × \frac{1}{2ϕ}$
$\displaystyle \sec 42° = \frac{4ϕ}{ϕ\sqrt{2+ϕ} \,+ \sqrt{3}}
= \frac{4ϕ\,(ϕ\sqrt{2+ϕ} \,- \sqrt{3})}{ϕ^2\,(2+ϕ)-3}
= ϕ\sqrt{2+ϕ} \,- \sqrt{3}$
First term: $\;ϕ\sqrt{2+ϕ} \,= \cot 18° = \tan 72°$
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