SOLUTION: Given the following equation: (sinx+cosx)^2= 1/2, Find x.

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Question 824883: Given the following equation: (sinx+cosx)^2= 1/2, Find x.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Given the following equation: (sinx+cosx)^2= 1/2, Find x.
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(sinx+cosx)^2= 1/2
sinx%2Bcosx=+sqrt%281%2F2%29
sin^2 + cos^2 = 1 --> cos = sqrt(1 - sin^2)
sin + sqrt(1 - sin^2) = sqrt(1/2)
Square both sides
sin^2 + 2*sin*sqrt(1 - sin^2) + (1 - sin^2) = 1/2
2*sin*sqrt(1 - sin^2) = -1/2
Square again
4sin^2(1 - sin^2) = 1/4
sin^2(1 - sin^2) = 1/16
sin^2 - sin^4 = 1/16
sin^4 - sin^2 + 1/16 = 0
Sub u for sin^2
u^2 - u + 1/16 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B0.0625+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A0.0625=0.75.

Discriminant d=0.75 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+0.75+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+0.75+%29%29%2F2%5C1+=+0.933012701892219
x%5B2%5D+=+%28-%28-1%29-sqrt%28+0.75+%29%29%2F2%5C1+=+0.0669872981077807

Quadratic expression 1x%5E2%2B-1x%2B0.0625 can be factored:
1x%5E2%2B-1x%2B0.0625+=+%28x-0.933012701892219%29%2A%28x-0.0669872981077807%29
Again, the answer is: 0.933012701892219, 0.0669872981077807. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B0.0625+%29

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sin^2 = 0.933012701892219
x = 75 degs
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sin^2 = 0.0669872981077807
x = 15 degs
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Those are the principal values, they're periodic
I'm not sure I did it the simplest way.