precalc math problem

Question

Solve the equation for exact solutions over the interval [0°,360°).

     

6sin(θ/2) =− 6cos(θ/2)

Answer 1

6sinT/2 =-6cosT/2

sinT/2 = -cosT/2

sinT/2)/cosT/2 = -1

tanT/2 = -1

T/2 = 135

T = 270 degrees

Answer 2

6sin(θ/2) = -6cos(θ/2)

6 √[(1 - cosθ)/2] = -6√[(1 + cosθ)/2]

Divide by 6 and then square both sides to get: 1 - cosθ = 1 + cosθ

2cosθ = 0

θ = 90° or 270°

90° is extraneous

So, θ = 270°