12cos^2θ + cosθ - 6 = 0 ? Quadratic type

how can I find values of θ in the interval 0 to 360 degrees


12cos^2θ + cosθ - 6 = 0 ?

Given 12cos^2(θ) + cos(θ) - 6 = 0 let x = cos(θ):

12x^2 + x - 6 = 0

Factor the quadratic:

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

Solve for x:

x = 2/3 or x = -3/4

Substitute x back into x = cos(θ):

cos(θ) = 2/3

or

cos(θ) = -3/4

cos(θ) = 2/3 when θ = 48.1896851 or when θ = 311.8103149

To find 311.8103149 just subtract the first answer (the answer the calculator would give you) from 360.

cos(θ) = -3/4 when θ = 138.5903779 or θ = 221.4096221.

Again to find 221.4096221 subtract the first answer, 138.5903779, from 360.

Let p=cosθ. Then 12p^2 + p - 6 = 0, so:
p = (-1 +/- squareroot(1^2-4(12)(-6)))/(2*12)
= (-1 +/- 17)/24 = 2/3 or -3/4

If p=cosθ = 2/3, then θ=cos^-1(2/3) or 360-cos^-1(2/3). Similarly for p=3/4. See this sketch to understand where each pair of solutions comes from: http://www.wolframalpha.com/input/?i=plo…

Treat cosθ as a variable for now; let cosθ = x

12x^2 + x - 6 = 0

I used an online calculator to solve for x:
x = -3/4, 2/3

cosθ = -3/4 or cosθ = 2/3.
θ = cos^-1(-3/4) or θ = cos^-1(2/3).

theta = 2 (pi n-tan^(-1)(1/sqrt(5), n element Z

~Kevin.

(3cosθ -2 )(4cosθ + 3) = 0
cosθ = ⅔ or cosθ = -¾
48.2°, 311.8° or 138.6°. 221.4°