Question 824883
Given the following equation: (sinx+cosx)^2= 1/2, Find x.
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(sinx+cosx)^2= 1/2
{{{sinx+cosx= sqrt(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
*[invoke solve_quadratic_equation 1,-1,1/16]
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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.