Asked by Sur
2sin^2x-1=0[0,2pi)
Answers
Answered by
Bosnian
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 )
sin ( x ) = 1 / sqrt ( 2 )
for
x = pi / 4 and x = 3 pi / 4
sin ( x ) = - 1 / sqrt ( 2 )
for
x = 5 pi / 4 and x = 7 pi / 4
2 sin ^ 2 ( x ) = 1 Divide both sides by 2
sin ^ 2 ( x ) = 1 / 2
sin ( x ) = ± 1 / sqrt ( 2 )
sin ( x ) = 1 / sqrt ( 2 )
for
x = pi / 4 and x = 3 pi / 4
sin ( x ) = - 1 / sqrt ( 2 )
for
x = 5 pi / 4 and x = 7 pi / 4
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