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

#1**+5 **

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