1. solve the equation for the interval [0, 2pie), cos^2x + 2 cos x +1=0?

Question

1. x= pie/4, 7pie/4
2. x=pie
3. x=pie/2, 3pie/2
4. x=2pie

Answer 1

0 = cos²x + 2cosx + 1
0 = (cosx + 1)²
0 = √(cosx + 1)²
0 = |cosx + 1|
0 = cosx + 1 (both roots)
cosx = -1
arccos(cosx) = arccos(-1)
x = arccos(-1)

On [0, 2π),

arccos(-1) occurs at π...answer (2).

Answer 2

Let cosx=t

t^2+2t+1=0

(t+1)^2=0

t=-1

cosx=-1

x=180 or pie

answer:
2. x=pie

Answer 3

(cosx+1)^2 = 0
cosx = -1
x = pi