The solution set of (5 + 4cos theta )(2cos th | vedclass

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(C) Given the equation: $(5 + 4\cos 	heta )(2\cos 	heta + 1) = 0$Case $1$: $5 + 4\cos 	heta = 0 \implies \cos 	heta = -\frac{5}{4}$.Since the rang

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$\left\{ \frac{2\pi }{3}, \frac{4\pi }{3} \right\}$

Vedclass · Answer

(C) Given the equation: $(5 + 4\cos 	heta )(2\cos 	heta + 1) = 0$Case $1$: $5 + 4\cos 	heta = 0 \implies \cos 	heta = -\frac{5}{4}$.Since the range of $\cos 	heta$ is $[-1, 1]$,this case has no real solution.Case $2$: $2\cos 	heta + 1 = 0 \implies \cos 	heta = -\frac{1}{2}$.In the interval $[0, 2\pi ]$,$\cos 	heta = -\frac{1}{2}$ at $	heta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $	heta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.Thus,the solution set is $\left\{ \frac{2\pi}{3}, \frac{4\pi}{3} ight\}$.

The solution set of $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$ in the interval $[0, 2\pi ]$ is

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