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

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

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

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

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

let y=sinx

12y^2 -y -6 =0

factor

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

y = 3/4, -2/3

sinx = 3/4, -2/3

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

use a calculator with an inverse sine function

if set on degrees, multiply by pi/180 to convert to radians in pi form

x = 45.6 degrees = .27pi, 134.4 degrees = .73pi, 221.8 degrees = 1.23pi, 318.2 degrees = 1.82pi,

x= about 45.5, 134.4, 221,8, 318.2 degrees

or x=about .27pi, .73pi, 1.23pi, 1.82pi radians (that's probably the form you want)

use pi = 3.141593 to convert to radians without pi if you want

x= about .79, 2.35, 3.87, 5.55 radians