Find all solutions on the interval [ 0 , 2 π ) . Give exact answers.

Question

 12cos^2(x)+sin(x)−6=0

Separate multiple solutions with a comma. If there are no solutions please enter, "no solution" exactly.

Answer 1

12cos²x + sinx − 6 = 0

12(1 - sin²x) + sinx - 6 = 0

12sin²x - sinx - 6 = 0

(4sinx - 3)(3sinx + 2) =0

sinx = 3/4 -> x = arcsin(3/4), π - arcsin(3/4) (the two angles on the unit circle with sinx = 3/4).

sinx = -2/3 -> x = π + arcsin(2/3), 2π - arcsin(2/3) (the two angles on the unit circle with sinx = -2/3)

Answer 2

12cos^2(x) + sinx -6 = 0 (use trig identity cos^2(x)+sin^2(x) = 1 and substitute)

12(1- sin^2(x)) + sinx -6 = 0

12 - 12sin^2(x) + sinx -6 = 0

-12sin^2(x) + sinx +6 = 0 (multiply by -1)

12sin^2(x) - sinx - 6 = 0

factor (if it helps, let y=sinx, convert to 12y^2 -y-6 =0, factor to (4y-3)(3y+2), y=3/4, -2/3)

(4sinx - 3)(3sinx +2) = 0

set each factor = 0

sinx= 3/4, sinx =-2/3

use a calculator with inverse trig functions

x = pi+arcsin(3/4), 2pi -arcsin(3/4), pi+ arcsin (-2/3), 2pi-sin(-2/3)

x =about 45.59 degrees, 134.41 degrees, 318.19 degrees, 221.81 degrees

multiply by pi/180 to convert to radians

x= about .27pi, .73pi, 1.77pi, 1.23pi