Trig Identity: Solving Cos(x/2) = 1/2 and the Proper Use of Plus or Minus

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The discussion centers on solving the equation Cos(x/2) = 1/2 and the proper application of the plus or minus sign in the square root. The participants clarify that using the positive root is not mandatory since Cos(x) can yield negative values in certain quadrants. They emphasize that solving for y first, where y = x/2, and then substituting back for x is a more effective approach. The confusion arises from the discrepancy between calculated answers and those provided in the textbook, particularly regarding the inclusion of additional solutions like 10pi/3. Ultimately, the importance of considering all possible values of x, including those from the arccos function, is highlighted.
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Cos(x/2) = 1/2

Trig Identity: Cos(x/2)= +- (1+Cosx/2)^(1/2) ..How do you know wheather to use the plus or minus of (1+Cosx/2)^(1/2) ? Do you only use the positive one because of the positive 1/2?
Anyways...

((1+Cosx)/2)^(1/2) = 1/2

(1+Cosx)/2 = 1/4

1+Cosx = 1/2

Cosx = -1/2

Ref Angle is pi/3 --> Cos is neg in Quad 2 and 3
Thus x Must be 2pi/3 and 4pi/3

However... in the back of the book it says the answers are 2pi/3 and 10pi/3...
How did I not come up with 10pi/3? what did I do wrong?
 
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This isn't the way you want to solve the problem. Let y=x/2, solve for y, then substitute back to solve for x.
 
Why don't you just Arccos both sides from the beginning? (I assume you capitalized Cos to mean the principal cosine)
 
Apphysicist said:
Why don't you just Arccos both sides from the beginning? (I assume you capitalized Cos to mean the principal cosine)

This will only give you one value of x. However, we want all possibilities for x.
 
Apphysicist and Mentallic are both right on this one. Taking the arccos first will give you two answers since cos is positive in two quadrants. Not quite sure what gb7nash is talking about.
 
Edit:

I just reread the original post. You're right. :smile:

If no info was given besides the equation though, arccos would only give you one value. To find all other values though:

{arccos + n*pi | n <- Z}
 
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