# PRECALC

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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"

• PRECALC -

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