Answers: 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}](/tpl/images/0147/6313/19c47.png)
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Solve on the interval (0,pi) 1+cosθ=((sqrt3)+2)/2...
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