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bridgetx516x
Group Title
solve the equation
cos2x−cosx=0
on the interval [0,2π)
 one year ago
 one year ago
bridgetx516x Group Title
solve the equation cos2x−cosx=0 on the interval [0,2π)
 one year ago
 one year ago

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ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

bridgetx516x Group TitleBest ResponseYou've already chosen the best response.0
i need to solve by factoring
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
cos 2x  cos x =0 cannot be factored. You could try a formula for cos 2x: cos 2x = 2cos²x1 If you use it, your equation becomes: 2cos²x1 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...
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
It should be soething like (2cos x +...)(cos x ...)=0
 one year ago

bridgetx516x Group TitleBest ResponseYou've already chosen the best response.0
oh i made a mistake while typing it.. it should be \[\cos^{2}xcosx=0\] can that be factored?
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
Yes, because the two terms have cos x in common. You can factor cosx, so it will be: cos x(.......  .....) = 0
 one year ago

getusel Group TitleBest ResponseYou've already chosen the best response.0
\[2\cos ^{2}xcosx1=0\] \[2\cos ^{2}x2cosx+cosx1=0\] (2cosx+1)(cosx1)=0 cosx=1/2,cosx=1 x=\[2\pi/3,0\]
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
@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.
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
@bridgetx516x: have you factored the equation in the mean time?
 one year ago

bridgetx516x Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
If you remember that cos²x is equal to (cos x)² = cos x * cos x, does that help maybe?
 one year ago

bridgetx516x Group TitleBest ResponseYou've already chosen the best response.0
so would it be cosx(cosx1)
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
Indeed! Even: cosx(cosx1)=0, so what would be your next move?
 one year ago

bridgetx516x Group TitleBest ResponseYou've already chosen the best response.0
split it in 2 equations. to get cosx=0 and cosx1=0 then the cosx=0 would be pi/2 and 3pi/2 cosx1=0 would now be cosx=1 the cosx=1 would just be 0 so the final answer should be.. 0,pi/2, 3pi/2
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
You're sounding like a pro here! Well done!
 one year ago

bridgetx516x Group TitleBest ResponseYou've already chosen the best response.0
thanks i mostly just got stuck on the factoring. thank you :)
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
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...)
 one year ago
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