SOLUTION: Solve the equation for 0≤x≤{{{2pi}}}. Write your answer as a multiple of {{{pi}}} if possible. {{{sin^2(x)}}}+{{{cos^2(-x)}}}= 2cos{{{(pi/2-(x))}}} Thanks!

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the equation for 0≤x≤{{{2pi}}}. Write your answer as a multiple of {{{pi}}} if possible. {{{sin^2(x)}}}+{{{cos^2(-x)}}}= 2cos{{{(pi/2-(x))}}} Thanks!      Log On


   

Question 1082992: Solve the equation for 0≤x≤2pi. Write your answer as a multiple of pi if possible.
sin%5E2%28x%29+cos%5E2%28-x%29= 2cos%28pi%2F2-%28x%29%29

Thanks!
Found 2 solutions by Gogonati, Alan3354:
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Sincecos%28-x%29=cos%28x%29 then sin%5E2%28x%29%2BCos%5E2%28-x%29=1%7D%7D%5D.+Also%7B%7B%7Bcos%28Pi%2F2-x%29=sin%28x%29 and the equation is written in the form:
2sin%28x%29=1 or sin%28x%29=1%2F2and its solutions are:
x=k%2A%28pi%29%2B%28pi%29%2F6,where k is an integer.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sin%5E2%28x%29%2B+cos%5E2%28-x%29=+2cos%28pi%2F2-%28x%29%29
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You're using more { } than necessary.
====================
cos(-x) = cos(x)
sin%5E2%28x%29%2B+cos%5E2%28x%29=+2cos%28pi%2F2-%28x%29%29
1+=+2cos%28pi%2F2-%28x%29%29
cos%28pi%2F2-%28x%29%29+=+1%2F2
sin%28x%29+=+1%2F2
x = pi/6, 5pi/6
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Write your answer as a multiple of pi if possible.
When is that not possible?