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Answer to Question #13063 in Trigonometry for Lucero

Question #13063
Expert's answer
2 sinx cosx+1-2sin2x=1
2 sinx cosx-2sin2x=0
2 sinx (cosx-sinx)=0
And we have two opportunities: either sinx=0 or cosx-sinx=0
1) sinx=0
x=pi n, n -integer
2) cosx-sinx=0
x=pi/4+2pi k, k-integer
So the answer is: pi n, pi/4+2pi k, where n,k-integers.

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