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)   (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

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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=0.75 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 0.933012701892219, 0.0669872981077807. Here's your graph:

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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.
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