Question

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

Answer 1

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