Question

Find the exact solution over the interval [0,2pie)
4cos ^2 x 8cos x +4=0

Answer 1

Is there a plus or a minus between 4cos^2(x) and 8cos(x) ?

Answer 2

a plus

Answer 3

4cos^2(x) + 8cos(x) + 4 = 0
Divide throughout by 4
cos^2(x) + 2cos(x) + 1 = 0
This is a quadratic equation in cos(x)
cos^2(x) + cos(x) + cos(x) + 1 = 0
cos(x)(cos(x) + 1) + 1(cos(x) + 1) = 0
(cos(x) + 1)(cos(x) + 1) = 0
(cos(x) + 1)^2 = 0
cos(x) + 1 = 0
cos(x) = -1
solve for x in the interval [0, 2pi)

Answer 4

cos(x) is minus 1 when x = pi.

Answer 5

thank you! now I understand