Please Help! Quadratic Work!

Question

Solve for x where 0 ≤ x ≤ 2pi.

cot^2 x + 2 cot x +1 = 0.

Answer 1

(cot x +1)(cot x + 1)=0

cot x = -1 => x = 135°

Answer 2

cot²x + 2cot x +1 = 0 This needs to be factored

(cot x+1)(cot x +1)=0 Now set each factor to zero and solve. Since they are the same, just need to do 1.

cot x+1=0 Subtract 1 from both sides

cot x = -1

The 2 values that give a cot of -1 between 0 and 2π are 3π/4 and 7π/4