Solving 2cot²x + 7cosec x - 13 = 0 for 0-360°

[Hootenanny]

The question is to solve the equation;
$$2\cot^{2}x + 7cosec x - 13 = 0$$
for all values of x between 0 and 360. I know I need to use a trig ident, but I've been using several over the course of an hour and I can't seem to get anywhere. Any hints would be helpful

 

[vaishakh]

Multiply with sin^2x

 

[Lyuokdea]

try using cot(x)^2 + 1 = csc(x)^2

~Lyuokdea

 

[Hootenanny]

Thank's got it.