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Sagot :
There are an infinite number of solutions, so I don't plan to list them all.
I'll list two of them, and then describe how to get all of the rest.
You said that 2cos(x) - 1 = 0
Add 1 to each side: 2cos(x) = 1
Divide each side by 2: cos(x) = 1/2
The angles whose cosine is 1/2 are 60 degrees, 300 degrees,
and any multiple of 360 degrees added to either of those.
I'll list two of them, and then describe how to get all of the rest.
You said that 2cos(x) - 1 = 0
Add 1 to each side: 2cos(x) = 1
Divide each side by 2: cos(x) = 1/2
The angles whose cosine is 1/2 are 60 degrees, 300 degrees,
and any multiple of 360 degrees added to either of those.
2cosx=1
cosx=1/2
x=cos∧-1(1/2)
x=60degrees or 360-60=300degrees
depending on your limits if in radians pi/3 and 5pi/3
cosx=1/2
x=cos∧-1(1/2)
x=60degrees or 360-60=300degrees
depending on your limits if in radians pi/3 and 5pi/3
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