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anonymous
 5 years ago
cos x/2=cos x. find all solutions in the interval [0,2pi).
anonymous
 5 years ago
cos x/2=cos x. find all solutions in the interval [0,2pi).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what angle and it's half have same cos value?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cos[0]1 Cos[pi/6]sqrt[3]/2 cos[pi/4]sqrt[2]/2 cos[pi/3]1/2 cos[pi/2]0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Huh? so, is that the answer? or...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(B/2) = ± sqrt([1 + cos B] / 2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos[x/2]=\[\pm \sqrt{1+\cos x \over 2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i put that in the interval as a possible solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{1+\cos(x) \over 2}= \cos(x)^2\] \[{1+\cos(x) }= \ 2cos(x)^2\] (2cos(x)+1)(cos(x)1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02cos(x)=1 arcos(cos(x))=arccos(1/2) x=2pi/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it would just be 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the solution would be x=0,4pi/3 x=0 1/2x=0 cos(0)= 1 cos(0)=1 x=4pi/3 1/2x=2pi/3 cos(4pi/3)=1/2 cos(2pi/3)=1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0than you. this helps a lot.
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