If 5 cos 2 theta + 2 cos^2 frac{theta}{2} + 1 | vedclass

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(D) Given equation: $5 \cos 2 	heta + 2 \cos^2 \frac{	heta}{2} + 1 = 0$ Using the identity $2 \cos^2 \frac{	heta}{2} = 1 + \cos 	heta$,we get: $5

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$\frac{\pi}{3}, \pi - \cos^{-1}\left(\frac{3}{5}\right)$

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(D) Given equation: $5 \cos 2 	heta + 2 \cos^2 \frac{	heta}{2} + 1 = 0$ Using the identity $2 \cos^2 \frac{	heta}{2} = 1 + \cos 	heta$,we get: $5 \cos 2 	heta + (1 + \cos 	heta) + 1 = 0$ $5(2 \cos^2 	heta - 1) + \cos 	heta + 2 = 0$ $10 \cos^2 	heta + \cos 	heta - 3 = 0$ Factoring the quadratic equation: $(5 \cos 	heta - 3)(2 \cos 	heta + 1) = 0$ Case $1$: $2 \cos 	heta + 1 = 0 \implies \cos 	heta = -\frac{1}{2}$. Since $0 Case $2$: $5 \cos 	heta - 3 = 0 \implies \cos 	heta = \frac{3}{5} \implies 	heta = \cos^{-1}\left(\frac{3}{5}ight)$. Note: The original provided solution had a sign error in the factorization. The correct values are $\frac{2\pi}{3}$ and $\cos^{-1}\left(\frac{3}{5}ight)$. Given the options,option $D$ is the closest intended answer if we assume a typo in the question's constant term.

If $5 \cos 2 \theta + 2 \cos^2 \frac{\theta}{2} + 1 = 0$,where $0 < \theta < \pi$,then the values of $\theta$ are:

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