SOLUTION: Find all values of "x" within the interval [0,2pi) for the equation: cos(x/2)= {{{ sqrt(2) }}}/2

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Question 106799This question is from textbook Precalculus with limits
: Find all values of "x" within the interval [0,2pi) for the equation:
cos(x/2)= +sqrt%282%29+/2
 This question is from textbook Precalculus with limits

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Since,
cos%28%28x%2F2%29%29=+sqrt%282%29%2F2
You know from the relationship,
sin%5E2%28y%29%2Bcos%5E2%28%28y%29%29=1
sin%5E2%28x%2F2%29%2B1%2F2=1
sin%28x%2F2%29=sqrt%282%29%2F2
As you can see on the unit circle, there are two angles (A and -A) that solve the equation,
cos%28%28x%2F2%29%29=+sqrt%282%29%2F2

You can determine A several ways.
Two ways are using the inverse trigonometric functions and geometrically.
Using inverse trig, the only angle that has equal sine and cosine is 45 degrees, or
A=45%5Eo
x%2F2=45%5Eo
x=90%5Eo
-A=45%5Eo
x%2F2=-45%5Eo
x=-90%5Eo
Geometrically, the triangle with A as one angle is a right tringle. It is also an right, isoceles triangle with a hypotneuse of 1 and sides of sqrt%282%29%2F2
Therefore A+A=90 or A=45.