Question

Find all solutions to the equation in the interval [0, 2π).

cos 2x - cos x = 0

0, 2pi/3, 4pi/3

pi/6, 5pi/6, 3pi/2

0, pi/2, 7pi/6, 11pi/6

No solution

Answer 1

Is the first term cos(two x) or cos-squared-x ?

Answer 2

cos 2x

Answer 3

Use the trig identity: cos(2x) = 2cos^2(x) - 1
Then it will be a quadratic equation in cos which can be solved

Answer 4

i don't know how to do tht

Answer 5

cos(2x) - cos(x) = 0
2cos^2(x) - 1 - cos(x) = 0
2cos^2(x) - cos(x) - 1= 0
Let u = cos(x)
2u^2 - u - 1 = 0
2u^2 - 2u + u - 1 = 0
2u(u-1) + 1(u-1) = 0
(2u+1)(u-1) = 0
u = -1/2 or u = 1
put back cos(x) in place of u:
cos(x) = -1/2 or cos(x) = 1
Find out all x in the interval [0,2pi] where cos(x) = -1/2 or 1

Answer 6

i don't understand what to do after that.....

Answer 7

@ranga

Answer 8

Find out all x in the interval [0,2pi] where cos(x) = -1/2 or 1
Look up the unit circle.
cos(x) is -1/2 when x = 2pi/3, 4pi/3
cos(x) is 1 when x = 0.

So x = 0, 2pi/3, 4pi/3