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bridgetx516x
 2 years ago
solve the equation
cos2x−cosx=0
on the interval [0,2π)
bridgetx516x
 2 years ago
solve the equation cos2x−cosx=0 on the interval [0,2π)

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ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2Write it as cos 2x = cos x and remember the general solution of cos a = cos b is a = b + 2kpi or a = b + 2kpi

bridgetx516x
 2 years ago
Best ResponseYou've already chosen the best response.0i need to solve by factoring

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2cos 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...

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2It should be soething like (2cos x +...)(cos x ...)=0

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

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2Yes, because the two terms have cos x in common. You can factor cosx, so it will be: cos x(.......  .....) = 0

getusel
 2 years ago
Best 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=\[2\pi/3,0\]

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2@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
 2 years ago
Best ResponseYou've already chosen the best response.2@bridgetx516x: have you factored the equation in the mean time?

bridgetx516x
 2 years ago
Best ResponseYou've already chosen the best response.0see 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
 2 years ago
Best ResponseYou've already chosen the best response.2If you remember that cos²x is equal to (cos x)² = cos x * cos x, does that help maybe?

bridgetx516x
 2 years ago
Best ResponseYou've already chosen the best response.0so would it be cosx(cosx1)

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2Indeed! Even: cosx(cosx1)=0, so what would be your next move?

bridgetx516x
 2 years ago
Best ResponseYou've already chosen the best response.0split 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

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2You're sounding like a pro here! Well done!

bridgetx516x
 2 years ago
Best ResponseYou've already chosen the best response.0thanks i mostly just got stuck on the factoring. thank you :)

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.2YW! 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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