Question

Solve cos² theta + costheta= 0 on the interval [0, 2π)

Answer 1

`\cos^2(\theta )+\cos(\theta )=0\implies \cos(\theta )( ~~ \cos(\theta )+1 ~~ )=0 \\\\[-0.35em] ~\dotfill\\\\ \cos(\theta )=0\implies \theta =\cos^(-1)(0)\implies \theta = \begin{cases} (\pi )/(2)\\[1em] (3\pi )/(2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \cos(\theta )+1=0\implies \cos(\theta )=-1\implies \theta =\cos^(-1)(-1)\implies \theta =\pi`

Answer 2

π/2, π, (3π)/2

Explanation:

cos²θ + cosθ = 0

let y = cosθ.

we have y² + y = 0.

use the quadratic formula:

y = ((-b ± √(b² - 4ac)) ÷ 2a)

= ((-1 ± √((1)² - 4(1)(0))) ÷ 2(1))

= (-1 ± 1) / 2

= 0 or -1.

so y = 0 or -1.

that means cosθ = 0 or -1.

cosθ = 0, θ = arcos 0 = π/2, (3π)/2

cosθ = -1, θ = arcos (-1) = π.

the solutions are π/2, π, (3π)/2.

see attachment for graph. the dots mark the solutions