anonymous
  • anonymous
cos x/2=cos x. find all solutions in the interval [0,2pi).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what angle and it's half have same cos value?
anonymous
  • anonymous
cos (x/2) = cos (x)
anonymous
  • anonymous
Cos[0]-1 Cos[pi/6]-sqrt[3]/2 cos[pi/4]-sqrt[2]/2 cos[pi/3]-1/2 cos[pi/2]-0

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anonymous
  • anonymous
Huh? so, is that the answer? or...
anonymous
  • anonymous
cos(B/2) = ± sqrt([1 + cos B] / 2)
anonymous
  • anonymous
cos[x/2]=\[\pm \sqrt{1+\cos x \over 2}\]
anonymous
  • anonymous
how do i put that in the interval as a possible solution?
anonymous
  • anonymous
[0,2pi)
anonymous
  • anonymous
\[{1+\cos(x) \over 2}= \cos(x)^2\] \[{1+\cos(x) }= \ 2cos(x)^2\] (2cos(x)+1)(cos(x)-1)
anonymous
  • anonymous
2cos(x)=-1 arcos(cos(x))=arccos(-1/2) x=2pi/3
anonymous
  • anonymous
So it would just be 0
anonymous
  • anonymous
the 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
  • anonymous
than you. this helps a lot.

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