Question

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

Answer 1

what angle and it's half have same cos value?

Answer 2

cos (x/2) = cos (x)

Answer 3

Cos[0]-1
Cos[pi/6]-sqrt[3]/2
cos[pi/4]-sqrt[2]/2
cos[pi/3]-1/2
cos[pi/2]-0

Answer 4

Huh? so, is that the answer? or...

Answer 5

cos(B/2) = ± sqrt([1 + cos B] / 2)

Answer 6

cos[x/2]=\[\pm \sqrt{1+\cos x \over 2}\]

Answer 7

how do i put that in the interval as a possible solution?

Answer 8

[0,2pi)

Answer 9

\[{1+\cos(x) \over 2}= \cos(x)^2\]
\[{1+\cos(x) }= \ 2cos(x)^2\]

(2cos(x)+1)(cos(x)-1)

Answer 10

2cos(x)=-1
arcos(cos(x))=arccos(-1/2)
x=2pi/3

Answer 11

So it would just be 0

Answer 12

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

Answer 13

than you. this helps a lot.