# Answer to Question #13063 in Trigonometry for Lucero

Question #13063

Sin2x+Cos2x=1

Expert's answer

2 sinx cosx+1-2sin

2 sinx cosx-2sin

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.

^{2}x=12 sinx cosx-2sin

^{2}x=02 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.

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