SOLUTION: solve 2cos2x-2cosx=0 over the interval [0,2pi)

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Question 1137706: solve 2cos2x-2cosx=0 over the interval [0,2pi)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

+2cos%282x%29-2cosx=0 ,[0,2pi) => means 0%3C=x%3C2pi

+2cos%282x%29-2cosx=0
2%28cos%282x%29-cosx%29=0

use identity
cos+%282x%29=2cos+%5E2%28x%29-1
+2%282cos+%5E2%28x%29-1-cos%28x%29%29=0
+2%282cos+%5E2%28x%29-cos%28x%29+-1%29=0.......if cos%28x%29=u
2%282u+%5E2-u+-1%29=0.....will be equal to zero if
%282u+%5E2-u+-1%29=0 ..........use quadratic formula



u=%28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F2a
u=%28-%28-1%29%2B-+sqrt%28%28-1%29%5E2-4%2A2%2A%28-1%29%29%29%2F%282%2A2%29
u=%281%2B-sqrt%281%2B8%29%29%2F4
u=%281%2B-sqrt%289%29%29%2F4
u=%281%2B-3%29%2F4
solutions:
u=4%2F4=1
u=-2%2F4=-1%2F2

since cos%28x%29=u, we have:
cos%28x%29=1 => since given interval 0%3C=x%3C2pi ,=> x=0
or cos%28x%29=-1%2F2=>x+=+2%2F3+%283pi%2A+n+%2B+pi%29 or x+=+2%2F3+%283pi%2A+n+-+pi%29, n element+Z
since given 0%3C=x%3C2pi =>x=2pi%2F3, x=4pi%2F3
so, your solutions are: x=0, x=2pi%2F3, and x=4pi%2F3