<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] $\begin{align} \cos(42°) &= \cos(60°-18°) \cr
&= \cos(60°)\cos(18°) + \sin(60°)\sin(18°) \cr
&= {1\over2} (\cos(18°) + \sqrt3 \sin(18°)) \cr
\sec(42°) &= \left({2 \over \cos(18°) + \sqrt3 \sin(18°)}\right) 
\left({\cos(18°) - \sqrt3 \sin(18°) \over \cos(18°) - \sqrt3 \sin(18°)}\right) \cr
&= {2(\cos(18°) - \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} \cr
&= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) - \sqrt3) \cr
\end{align}$ 
Let $s=\sin(18°)$, using multiple angles formula
$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s = 1$ 
$16s^5 - 20s^3 + 5s - 1 = 0$
$(s-1)(4s^2+2s-1)^2 = 0$
Since $s≠1$, we get $4s^2+2s-1 = 0\quad → \large{2s \over 1-4s^2} = 1$
$\tan(x) = \sec(42°) + \sqrt3 = (\cot(18°) - \sqrt3) + \sqrt3 = \tan(72°)$
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