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# Find all real numbers in the interval [0,2π]

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Find all numbers in the interval [0,2π] that staisfy the equation. Round solutions to the nearest tenth as needed. State rounded soulutions in soulution set form.

27sin^2x-3sinx=cos^2x

Not really sure where to start with this, any help would be appreciated.

Guest Mar 11, 2017
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27sin^2x-3sinx=cos^2x

27sin^2x - 3sin x - cos^2x = 0

27sin^2x - 3 sinx - (1 - sin^2x) = 0

28sin^2x - 3sin x - 1   = 0   factor as

(7sin x + 1) (4sin x - 1)  = 0

Set each factor to 0  and we have

7 sin x + 1   = 0     subtract 1 from both sides

7sin x  = -1    divide both sides by 7

sinx = (-1/7)     take the sine inverse

arctan (-1/7) = x  ≈ -8.2°  ≈ 351.8°  and ≈ 188.2°

4sin x - 1  = 0    add 1 to both sides

4sinx = 1    divide both sides by 4

sin x = 1/4 take the sine inverse

arcsin (1/4) = x  ≈ 14.5°  and ≈ 165.5°

Solutions [ 14.5°, 165.5°, 188.2°, 351.8° ]

CPhill  Mar 11, 2017

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