How do you solve cos^2 x - 4 cos x = 1 over the interval 0 to 2pi?

1 Answer
Nghi N.
Apr 5, 2016

103^@65 and 256^@35

Explanation:

f(x) = cos^2 x - 4cos x - 1 = 0
D = d^2 = b^2 - 4ac = 16 + 4 = 20 ---> d = +- 2sqrt5
There are 2 real roots.
cos x = -b/(2a) +- d/(2a) = 4/2 +- 2sqrt5/2 = 2 +- sqrt5

a . cos x = 2 + sqrt5 (Rejected because > 1)
b. cos x = 2 - sqrt5 = 2 - 2.24 = - 0. 24 -->x = +- 103^@65
Answer for (0, 2pi):
103^@65 and 256^@35 (co-terminal to - 103.65)

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
4202 views around the world
You can reuse this answer
Creative Commons License