Question

The no of possible value of x (0,2pi) which satisfy the eq cotx-cosecx= 2sinx is

Answer 1

$\frac{ \sin( \alpha ) }{ \ \sin) ( \alpha ) }$
$\frac{ \ \cos ( \alpha ) }{ \ \sin( \alpha ) } - \frac{1}{ \sin( \alpha ) } = 2 \sin( \alpha ) \\ \cos( \alpha ) - 1 = 2 { \sin }^{2} \alpha \\ \cos( \alpha ) - 1 = 2 - 2 { \cos}^{2} \alpha \\ 2 { \cos }^{2} \alpha + \cos( \alpha ) - 3 = 0 \\ let \: \cos( \alpha ) = y \\ 2 {y}^{2} + y - 3 = 0 \\ 2 {y}^{2} + 3y - 2y - 3 = 0 \\ 2y(y - 1) + 3(y - 1) = 0 \\ (2y + 3)(y - 1) = 0 \\ y = - \frac{3}{2} ..not \: possible. \\ y = 1 \\ \\ \\ in \: (0 \: \: 2\pi) \\ cos \alpha = 90 \: at \: (0) \: and \: 2\pi$
but there points are excluded as given in question hence zero solutions