The solution set of 8 cos^2 theta + 14 cos th | vedclass

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(C) Given equation: $8 \cos^2 	heta + 14 \cos 	heta + 5 = 0$ Factorizing the quadratic equation: $8 \cos^2 	heta + 10 \cos 	heta + 4 \cos 	heta +

Vedclass · Accepted Answer

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

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(C) Given equation: $8 \cos^2 	heta + 14 \cos 	heta + 5 = 0$ Factorizing the quadratic equation: $8 \cos^2 	heta + 10 \cos 	heta + 4 \cos 	heta + 5 = 0$ $	herefore 2 \cos 	heta (4 \cos 	heta + 5) + 1 (4 \cos 	heta + 5) = 0$ $	herefore (2 \cos 	heta + 1)(4 \cos 	heta + 5) = 0$ $	herefore \cos 	heta = -\frac{1}{2}$ or $\cos 	heta = -\frac{5}{4}$ Since the range of $\cos 	heta$ is $[-1, 1]$,the value $\cos 	heta = -\frac{5}{4}$ is impossible. $	herefore \cos 	heta = -\frac{1}{2}$ In the interval $[0, 2\pi]$,$\cos 	heta = -\frac{1}{2}$ at $	heta = \frac{2\pi}{3}$ and $	heta = \frac{4\pi}{3}$. Thus,the solution set is $\left\{\frac{2\pi}{3}, \frac{4\pi}{3}ight\}$.

The solution set of $8 \cos^2 \theta + 14 \cos \theta + 5 = 0$ in the interval $[0, 2\pi]$ is

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