If 5cos 2theta + 2cos^2frac{theta}{2} + 1 = 0 | vedclass

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(D) Given the equation: $5\cos 2	heta + 2\cos^2\frac{	heta}{2} + 1 = 0$Using the identities $\cos 2	heta = 2\cos^2	heta - 1$ and $2\cos^2\frac{	h

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

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(D) Given the equation: $5\cos 2	heta + 2\cos^2\frac{	heta}{2} + 1 = 0$Using the identities $\cos 2	heta = 2\cos^2	heta - 1$ and $2\cos^2\frac{	heta}{2} = 1 + \cos	heta$,we get:$5(2\cos^2	heta - 1) + (1 + \cos	heta) + 1 = 0$$10\cos^2	heta - 5 + 1 + \cos	heta + 1 = 0$$10\cos^2	heta + \cos	heta - 3 = 0$Factoring the quadratic equation:$(5\cos	heta - 3)(2\cos	heta + 1) = 0$This gives $\cos	heta = \frac{3}{5}$ or $\cos	heta = -\frac{1}{2}$.For $\cos	heta = \frac{3}{5}$,$	heta = \pm\cos^{-1}\frac{3}{5}$.For $\cos	heta = -\frac{1}{2}$,$	heta = \pm\frac{2\pi}{3}$.Since the provided options are limited,the correct values within the range $(-\pi, \pi)$ are $\pm\frac{2\pi}{3}$ and $\pm\cos^{-1}\frac{3}{5}$.

If $5\cos 2\theta + 2\cos^2\frac{\theta}{2} + 1 = 0$ and $-\pi < \theta < \pi$,then $\theta = $

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