How do you solve \sin^2x-\cos^2x=0 for x in the interval

kerrum75 2021-12-13 Answered
How do you solve sin2xcos2x=0 for x in the interval [0,2π) ?
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usumbiix
Answered 2021-12-14 Author has 33 answers
Applying the identity cos2(x)sin2(x)=cos(2x) we have
sin2(x)cos2(x)=cos(2x)
In general,
cos(u)=0u=nπ2 for some nZ
Thus we have
sin2(x)cos2(x)=0
2x=nπ2 for nZ
x=nπ4 for nZ
Restricting our values to the interval [0,2π] gives our final result:
x(π4,3π4,5π4,7π4)
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lovagwb
Answered 2021-12-15 Author has 50 answers

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