Please Help! (Quadratics)

Question

The problem says to solve for x where 0 ≤ x ≤ 2pi.

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

It also says that the answers should be given as fractions?

Answer 1

Can you visualize how to solve U² + 2U + 1 = 0? This is a perfect square trinomial which can be written as :

(U + 1)² = 0. You could also solve for U by using the quadratic formula.

The above results in U = -1. A root with multiplicity 2.

Now instead of U, visualize U replaced with cot x. Now we end up with (cot x +1)² = 0, leading to"

cot x = -1.

Solving for x, x = arc cot (-1). Between o and 2pi, what angles have a cotangent of -1?

Cotangent is negative in 2nd and 4th quadrants. The reference angle is pi/4 (45 degrees). Leave it to you to determine the angles in 2nd and 4th quadrants with ref angle of pi/4.