Asked by Mark
Solve the equation sin^2x-cos^2x=0 over the interval [0, 2pi). Having trouble getting started.
Answers
Answered by
Bosnian
cos ^ 2 ( x ) = 1 - sin ^ 2 ( x )
sin ^ 2 ( x ) - cos ^ 2 ( x ) = 0
sin ^ 2 ( x ) - [ 1 - sin ^ 2 ( x ) ] = 0
sin ^ 2 ( x ) - 1 + sin ^ 2 ( x ) = 0
2 sin ^ 2 ( x ) - 1 = 0
2 sin ^ 2 ( x ) = 1 Divide both sides by 2
sin ^ 2 ( x ) = 1 / 2
sin ( x ) = ± 1 / sqrt ( 2 )
Solutions :
x = pi / 4
[ pi / 4 = 45 ° , sin ( 45 ° ) = 1 / sqrt ( 2 ) ]
x = 3 pi / 4
[ 3 pi / 4 = 135 ° , sin ( 135 ° ) = 1 / sqrt ( 2 ) ]
x = 5 pi / 4
[ 5 pi / 4 = 225 ° , sin ( 225 ° ) = - 1 / sqrt ( 2 ) ]
x = 7 pi / 4
[ 7 pi / 4 = 315 ° , sin ( 315 ° ) = - 1 / sqrt ( 2 ) ]
sin ^ 2 ( x ) - cos ^ 2 ( x ) = 0
sin ^ 2 ( x ) - [ 1 - sin ^ 2 ( x ) ] = 0
sin ^ 2 ( x ) - 1 + sin ^ 2 ( x ) = 0
2 sin ^ 2 ( x ) - 1 = 0
2 sin ^ 2 ( x ) = 1 Divide both sides by 2
sin ^ 2 ( x ) = 1 / 2
sin ( x ) = ± 1 / sqrt ( 2 )
Solutions :
x = pi / 4
[ pi / 4 = 45 ° , sin ( 45 ° ) = 1 / sqrt ( 2 ) ]
x = 3 pi / 4
[ 3 pi / 4 = 135 ° , sin ( 135 ° ) = 1 / sqrt ( 2 ) ]
x = 5 pi / 4
[ 5 pi / 4 = 225 ° , sin ( 225 ° ) = - 1 / sqrt ( 2 ) ]
x = 7 pi / 4
[ 7 pi / 4 = 315 ° , sin ( 315 ° ) = - 1 / sqrt ( 2 ) ]
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