Answer:
[tex]\theta =\dfrac{2}{3} \pi ,\;\; \theta =\dfrac{4}{3} \pi[/tex]
Step-by-step explanation:
Given:
[tex]\cos 2\theta=-5 \cos \theta-3, \quad \quad 0 \leq \theta < 2 \pi[/tex]
[tex]\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}}[/tex]
Use the cos double angle identity to create a quadratic:
[tex]\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}[/tex]
Factor the quadratic:
[tex]\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}[/tex]
Apply the zero-product property:
[tex]2 \cos \theta + 1 = 0 \implies \cos \theta =-\dfrac{1}{2}[/tex]
[tex]\cos \theta + 2= 0 \implies \cos \theta =-2[/tex]
As -1 ≤ cos θ ≤ 1, cos θ = -2 is undefined.
Therefore, the only valid solution is cos θ = -¹/₂
Therefore:
[tex]\implies \cos \theta =-\dfrac{1}{2}[/tex]
[tex]\implies \theta =\cos^{-1}\left(-\dfrac{1}{2}\right)[/tex]
[tex]\implies \theta =\dfrac{2}{3} \pi +2\pi n,\;\; \dfrac{4}{3} \pi+2\pi n[/tex]
Solutions in the given interval 0 ≤ θ < 2π :
[tex]\theta =\dfrac{2}{3} \pi ,\;\; \theta =\dfrac{4}{3} \pi[/tex]
