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- anonymous

solve the equation
cos2x−cosx=0
on the interval [0,2π)

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- anonymous

solve the equation
cos2x−cosx=0
on the interval [0,2π)

- jamiebookeater

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- ZeHanz

Write it as cos 2x = cos x and remember the general solution of
cos a = cos b
is a = b + 2kpi or a = -b + 2kpi

- anonymous

i need to solve by factoring

- ZeHanz

cos 2x - cos x =0 cannot be factored.
You could try a formula for cos 2x:
cos 2x = 2cos²x-1
If you use it, your equation becomes:
2cos²x-1 -cosx=0 or better: 2cos²x - cos x - 1 = 0.
This is a 2nd degree equation with cos x as a variable.
Try to factor it...

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- ZeHanz

It should be soething like (2cos x +...)(cos x -...)=0

- anonymous

oh i made a mistake while typing it.. it should be \[\cos^{2}x-cosx=0\]
can that be factored?

- ZeHanz

Yes, because the two terms have cos x in common. You can factor cosx, so it will be:
cos x(....... - .....) = 0

- anonymous

\[2\cos ^{2}x-cosx-1=0\]
\[2\cos ^{2}x-2cosx+cosx-1=0\]
(2cosx+1)(cosx-1)=0
cosx=-1/2,cosx=1
x=\[2\pi/3,0\]

- ZeHanz

@getusel: the equation has changed in the mean time :(
Also, I think it is better not to give the complete calculation.
Giving hints to the solution is better.

- ZeHanz

@bridgetx516x: have you factored the equation in the mean time?

- anonymous

see this is where i originally got stuck because it will be cos(___-1) but i am not sure what to put in the ___ because of the square

- ZeHanz

If you remember that cos²x is equal to (cos x)² = cos x * cos x, does that help maybe?

- anonymous

so would it be cosx(cosx-1)

- ZeHanz

Indeed!
Even: cosx(cosx-1)=0, so what would be your next move?

- anonymous

split it in 2 equations. to get cosx=0 and cosx-1=0
then the cosx=0 would be pi/2 and 3pi/2
cosx-1=0 would now be cosx=1
the cosx=1 would just be 0
so the final answer should be..
0,pi/2, 3pi/2

- ZeHanz

You're sounding like a pro here!
Well done!

- anonymous

thanks i mostly just got stuck on the factoring.
thank you :)

- ZeHanz

YW!
BTW, I am surprised that even your misspelled first equation could be done with factoring!
(although that would be a little over the top, because there would be an easier way...)

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