Solve the equation sin^2x-cos^2x=0 over the interval [0, 2pi). Having trouble getting started.

Answer 1

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 ) ]