A geometry problem by Benjamin ononogbu

$\cos \theta = 3$ does not have a real solution but a complex one.

$\begin{aligned} \cos \theta & = 3 \\ \Rightarrow \frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right) & = 3 \\ e^{i\theta} + e^{-i\theta} & = 6 \\ e^{i2\theta} - 6 e^{i\theta} + 1 & = 0 \quad \quad \quad \quad \quad \quad \space \space \small \color{#3D99F6}{\text{A quadratic of }e^{i \theta}} \\ \Rightarrow e^{i\theta} & = \frac{6 \pm \sqrt{36-4}}{2} \\ & = 3 \pm 2 \sqrt{2} \\ i\theta & = \ln (3 \pm 2 \sqrt{2}) \quad \quad \small \color{#3D99F6}{\text{Squaring both sides,}} \\ -\theta^2 & = \begin{cases} \ln^2 (3 + 2 \sqrt{2}) = 1.762747174^2 & = 3.1072776 \\ \ln^2 (3 - 2 \sqrt{2}) = (-1.762747174)^2 &=3.1072776 \end{cases} \\ \Rightarrow \theta^2 & = \boxed{-3.107} \end{aligned}$

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$\color{#3D99F6}{\text{Solution suggested by }}$ Shyamdhu Mukherjee :

$\cos \theta = 3 \quad \Rightarrow \sin \theta = \sqrt{1-3^2} = i2\sqrt{2}$ .

Therefore,

$\begin{aligned} e^{i \theta} & = \cos \theta + i \sin \theta \\ & = 3 \pm i (i 2\sqrt{2}) \\ & = 3 \pm 2\sqrt{2} \\ \Rightarrow i \theta & = \ln (3 \pm 2\sqrt{2}) \\ \theta^2 & = \ln^2 (3 \pm 2\sqrt{2}) \\ & = \boxed{-3.107} \end{aligned}$