Question

Solve the equation cos 2 θ = -5 cos θ - 3 on the interval 0 ≤ θ < 2π

Answers must be exact.

Answer 1

`\theta =(2)/(3) \pi ,\;\; \theta =(4)/(3) \pi`

Explanation:

Given:

`\cos 2\theta=-5 \cos \theta-3, \quad \quad 0 \leq \theta < 2 \pi`

`\boxed{\begin{minipage}{6.5 cm}\underline{Cos Double Angle Identity}\\\\$\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B$\\\end{minipage}}`

Use the cos double angle identity to create a quadratic:

`\begin{aligned}\cos 2\theta &=-5 \cos \theta-3\\\cos^2\theta - \sin^2 \theta &=-5 \cos \theta-3\\\cos^2\theta - (1-\cos^2 \theta) &=-5 \cos \theta-3\\2\cos^2\theta - 1 &=-5 \cos \theta-3\\2\cos^2\theta+5 \cos \theta +2 &=0\end{aligned}`

Factor the quadratic:

`\begin{aligned}2\cos^2\theta+5 \cos \theta +2 &=0\\2\cos^2\theta+4 \cos \theta +\cos \theta +2 &=0\\2\cos \theta(\cos \theta +2)+1(\cos \theta +2) &=0\\(2\cos \theta+1)(\cos \theta +2)&=0\end{aligned}`

Apply the zero-product property:

`2 \cos \theta + 1 = 0 \implies \cos \theta =-(1)/(2)`

`\cos \theta + 2= 0 \implies \cos \theta =-2`

As -1 ≤ cos θ ≤ 1, cos θ = -2 is undefined.

Therefore, the only valid solution is cos θ = -¹/₂

Therefore:

`\implies \cos \theta =-(1)/(2)`

`\implies \theta =\cos^(-1)\left(-(1)/(2)\right)`

`\implies \theta =(2)/(3) \pi +2\pi n,\;\; (4)/(3) \pi+2\pi n`

Solutions in the given interval 0 ≤ θ < 2π :

`\theta =(2)/(3) \pi ,\;\; \theta =(4)/(3) \pi`