SOLUTION: 7) For x E [0,2pi), solve: cos2x = cosx a) 2 pi/3, 4pi/3 and pi b) 2 pi/3, 4 pi/3 and 0 c) Pi/3, 5pi/3 and 0 d) Pi/6, 5pi/6 and pi e)

Algebra ->  Trigonometry-basics -> SOLUTION: 7) For x E [0,2pi), solve: cos2x = cosx a) 2 pi/3, 4pi/3 and pi b) 2 pi/3, 4 pi/3 and 0 c) Pi/3, 5pi/3 and 0 d) Pi/6, 5pi/6 and pi e)       Log On


   

Question 121956: 7) For x E [0,2pi), solve: cos2x = cosx
a) 2 pi/3, 4pi/3 and pi
b) 2 pi/3, 4 pi/3 and 0
c) Pi/3, 5pi/3 and 0
d) Pi/6, 5pi/6 and pi
e) 5pi/6, 7pi/6 and pi
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given equation

Replace with . Note: I'm using the identity

Subtract cos%28x%29 from both sides

Rearrange the terms

Now let u=cos%28x%29

Replace each cos%28x%29 with u

%28u-1%29%282u%2B1%29=0 Factor the left side

Now set each factor equal to zero

u-1=0 or 2u%2B1=0

Now solve for u in each case

u=1 or u=-1%2F2

Now remember we let u=cos%28x%29. So this means

cos%28x%29=1 or cos%28x%29=-1%2F2

So let's solve cos%28x%29=1 to get x=0 or x=2pi. However, since 2pi is excluded the only solution for cos%28x%29=1 is x=0

Now let's solve cos%28x%29=-1%2F2 to get x=2pi%2F3 or x=4pi%2F3

So putting these solutions together, we get:

x=0, x=2pi%2F3 or x=4pi%2F3


So this means the answer is B)