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ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1Write 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
 one year ago
Best ResponseYou've already chosen the best response.0i need to solve by factoring

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

bridgetx516x
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Yes, because the two terms have cos x in common. You can factor cosx, so it will be: cos x(.......  .....) = 0

getusel
 one year 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
 one year ago
Best 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.

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

bridgetx516x
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1If you remember that cos²x is equal to (cos x)² = cos x * cos x, does that help maybe?

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

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

bridgetx516x
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1You're sounding like a pro here! Well done!

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

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1YW! 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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