Find all solutions of each equation in the interval [0,2π). 2 cos2Θ + 3 cos Θ = 2

Question

Find all solutions of each equation in the interval [0,2π).

2 cos²Θ + 3 cos Θ = 2

Please say why some solutions will work and some will not.

Thank you!

Answer 1

x=pi/3, 5pi/3, 60 or 300 degrees

let x=cosT=cos(theta), then

2x^2 +3x -2=0

(2x-1)(x+2)=0

x=1/2=cosT, T=arccos5= pi/3 or 5pi/3

x=-2=cosT =undefined, as no cosines are <-1

Answer 2

2cos²(θ) + 3cos(θ) = 2

Let u = cos(θ)

2u² + 3u - 2 = 0

u = (1/(2*2))[ -3 ± √( 3² - 4(2)(-2) ) ]

u = (1/4)[-3 ± √(9 + 16)]

u = (1/4)[-3 ± 5]

u = -8/4 , 2/4 ==> u = -2, 1/2

u = cos(θ) = -2, 1/2

The range of both sin() and cos() are [-1, 1], so we rule out cos() = -2

==> cos(θ) = 1/2

==> θ = ± 60º = ± π/3 in Quadrants I & IV

π/3 is fine, -π/3 is also valid, but we need to express it in the requested range [0, 2π]

2π - π/3 = 6π/3 - π/3 = 5π / 3

Ans. θ = π / 3, 5π / 3

Answer 3

This equation is quadratic in terms of cos(θ), so set it equal to zero, then solve that quadratic equation by factoring.

2cos²(θ) + 3cos(θ) - 2 = 0

2cos²(θ) + 4cos(θ) - 1cos(θ) - 2 = 0 (slit-the-middle)

2cos(θ)[cos(θ) + 2] - 1[cos(θ) + 2] = 0 (factor GCF from 1st and 2nd groups)

[cos(θ) + 2][2cos(θ) -1] = 0 [factor out common binomial factor)

cos(θ) + 2 = 0 OR 2cos(θ) - 1 = 0 {zero product property)

For 1st equation:

cos(θ) = -2

θ = cos^-1(-2)

θ = {} (cosine values must be between -1 and 1 inclusive)

For 2nd equation:

2cos(θ) = 1

cos(θ) = 1/2

θ = cos^-1(1/2)

θ= π/3 or 5π/3 (cosine positive in 1st and 4th quadrants--reference angle π/3)

Here is a way to check the results on Desmos:

desmos.com/calculator/qcie5dsd8n