Another Trig. question

Q. Use an algebraic method to find the solution for 0 < x < 2pi of the equation 5cot x + 2 cosec^2 x = 5.
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I worked out the above and got cot x = 1/2 & -3... Not sure where to go from here .

sin²x + cos²x = 1
1 + cot²x = cosec²x

=> 5cotx + 2cosec²x = 5
= 5cotx + 2(1+cot²x) = 5
5cotx + 2 + 2cot²x = 5
2cot²x + 5cotx - 3 = 0
(2cotx-1)(cotx+3) = 0
cotx = ½ OR cotx = -3
1/tanx = ½ OR 1/tanx = -3
tanx = 1/½ = 2
tanx = 1/-3 = -1/3

x = arctan(2) OR x = arctan(-1/3)