Question

2csc2x - 2cscx - 1 = 0 for 0° ≤ x < 360°

Answer 1

You have a quadratic equation.
\[ay^2+by+c=0\]
The solutions are
\[y=(-b \pm \sqrt{b^2-4ac})/(2a)\]
Then,
\[\csc x =1/sinx\]