PRECALC
posted by greeny .
solve for x in the interval zero to two pi.
2 sin 4x = 1
i got 5 pi/24 and pi/ 24 but my teacher said that there should be eight solutions. what would those be and how would i find it"

2 sin 4 x = 1 Divide both sides by 2
sin 4 x = 1 / 2
Take the inverse sine of both sides :
4 x = sin ^  1 ( 1 / 2 )
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Remark :
sin ^  1 ( 1 / 2 ) =
pi / 6 + 2 pi n
OR
5 pi / 6 + 2 pi n
Where n is an integer
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4 x = pi / 6 + 2 pi n
OR
4 x = 5 pi / 6 + 2 pi n
Divide both sides by 4
x = pi / 24 + pi n / 2
OR
x = 5 pi / 24 + pi n / 2
For n = 0
x = pi / 24 + pi * 0 / 2
x = pi / 24
OR
x = 5 pi / 24 + pi * 0 / 2
x = 5 pi / 24
For n = 1
x = pi / 24 + pi * 1 / 2
x = pi / 24 + pi / 2
x = pi / 24 + 12 pi / 24
x = 13 pi / 24
OR
x = 5 pi / 24 + pi * 1 / 2
x = 5 pi / 24 + pi / 2
x = 5 pi / 24 + 12 pi / 24
x = 17 pi / 24
For n = 2
x = pi / 24 + pi * 2 / 2
x = pi / 24 + pi
x = pi / 24 + 24 pi / 24
x = 25 pi / 24
OR
x = 5 pi / 24 + pi * 2 / 2
x = 5 pi / 24 + pi
x = 5 pi / 24 + 24 pi / 24
x = 29 pi / 24
For n = 3
x = pi / 24 + pi * 3 / 2
x = pi / 24 + 3 pi / 2
x = pi / 24 + 3 * 12 pi / 24
x = pi / 24 + 36 pi / 24
x = 37 pi / 24
OR
x = 5 pi / 24 + pi * 3 / 2
x = 5 pi / 24 + 3 pi / 2
x = 5 pi / 24 + 3 * 12 pi / 24
x = 5 pi / 24 + 36 pi / 24
x = 41 pi / 24
Solutions :
x1 = pi / 24
x2 = 5 pi / 24
x3 = 13 pi / 24
x4 = 17 pi / 24
x5 = 25 pi / 24
x6 = 29 pi / 24
x7 = 37 pi / 24
x8 = 41 pi / 24