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

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(C) Given the equation: $(5+4 \cos 	heta)(2 \cos 	heta+1)=0$. Since $5+4 \cos 	heta$ can never be $0$ (because $-1 \le \cos 	heta \le 1$,so $1 \le

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

Vedclass · Answer

(C) Given the equation: $(5+4 \cos 	heta)(2 \cos 	heta+1)=0$. Since $5+4 \cos 	heta$ can never be $0$ (because $-1 \le \cos 	heta \le 1$,so $1 \le 5+4 \cos 	heta \le 9$),we must have $2 \cos 	heta + 1 = 0$. $\Rightarrow \cos 	heta = -\frac{1}{2}$. In the interval $[0, 2 \pi]$,$\cos 	heta$ is negative in the second and third quadrants. $\cos 	heta = \cos \left(\pi - \frac{\pi}{3}ight) = \cos \left(\frac{2 \pi}{3}ight)$ and $\cos 	heta = \cos \left(\pi + \frac{\pi}{3}ight) = \cos \left(\frac{4 \pi}{3}ight)$. 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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