Question

2cos^2x+sinx=2 domain restriction 0,2pi

Answer 1

$2\cos ^{ 2 }{ x } +\sin { x } =2\\ \\ 2\left( 1-\sin ^{ 2 }{ x } \right) +\sin { x } =2\\ \\ 2-2\sin ^{ 2 }{ x } +\sin { x } =2$

$\\ \\ 2\sin ^{ 2 }{ x } -\sin { x } +2-2=0\\ \\ 2\sin ^{ 2 }{ x } -\sin { x } =0\\ \\ \sin { x } \left( 2\sin { x } -1 \right) =0$

$\therefore \quad \sin { x } =0\\ \\ \therefore \quad x=0,\quad \pi ,\quad 2\pi$

$\therefore \quad 2\sin { x } -1=0\\ \\ 2\sin { x } =1\\ \\ \sin { x } =\frac { 1 }{ 2 } \\ \\ \therefore \quad x=\frac { \pi }{ 6 } ,\quad \frac { 5\pi }{ 6 }$