Answer to Question #13063 in Trigonometry for Lucero

Question

Sin2x+Cos2x=1

Answer 1

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
tanx=1
x=pi/4+2pi k, k-integer
So the answer is: pi n, pi/4+2pi k, where n,k-integers.