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

Trigonometry
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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
i need to solve by factoring
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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Other answers:

It should be soething like (2cos x +...)(cos x -...)=0
oh i made a mistake while typing it.. it should be \[\cos^{2}x-cosx=0\] can that be factored?
Yes, because the two terms have cos x in common. You can factor cosx, so it will be: cos x(....... - .....) = 0
\[2\cos ^{2}x-cosx-1=0\] \[2\cos ^{2}x-2cosx+cosx-1=0\] (2cosx+1)(cosx-1)=0 cosx=-1/2,cosx=1 x=\[2\pi/3,0\]
@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.
@bridgetx516x: have you factored the equation in the mean time?
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
If you remember that cos²x is equal to (cos x)² = cos x * cos x, does that help maybe?
so would it be cosx(cosx-1)
Indeed! Even: cosx(cosx-1)=0, so what would be your next move?
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
You're sounding like a pro here! Well done!
thanks i mostly just got stuck on the factoring. thank you :)
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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