Question

Solve on the interval (0,pi) 1+cosθ=((sqrt3)+2)/2

Answer 1

$\bf 1+cos(\theta )=\cfrac{\sqrt{3}+2}{2}\implies cos(\theta )=\cfrac{\sqrt{3}+2}{2}-1 \\\\\\ cos(\theta )=\cfrac{\sqrt{3}+2}{2}-\cfrac{2}{2}\implies cos(\theta )=\cfrac{\sqrt{3}\underline{+2-2}}{2}\implies cos(\theta )=\pm\cfrac{\sqrt{3}}{2} \\\\\\ cos^{-1}[cos(\theta )]=cos^{-1}\left( \pm\cfrac{\sqrt{3}}{2} \right)\implies \measuredangle \theta = \begin{cases} \frac{\pi }{6}\\\\ \frac{5\pi }{6}\\\\ \frac{7\pi }{6}\\\\ \frac{11\pi }{6} \end{cases}$