If cos theta + sec theta = frac{5}{2},then th | vedclass

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(D) Given the equation: $\cos 	heta + \sec 	heta = \frac{5}{2}$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we have $\cos 	heta + \frac{1}{\cos 	h

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$2n\pi \pm \frac{\pi}{3}$

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(D) Given the equation: $\cos 	heta + \sec 	heta = \frac{5}{2}$.Since $\sec 	heta = \frac{1}{\cos 	heta}$,we have $\cos 	heta + \frac{1}{\cos 	heta} = \frac{5}{2}$.Multiplying by $2 \cos 	heta$,we get $2 \cos^2 	heta + 2 = 5 \cos 	heta$,which simplifies to $2 \cos^2 	heta - 5 \cos 	heta + 2 = 0$.Factoring the quadratic equation: $(2 \cos 	heta - 1)(\cos 	heta - 2) = 0$.This gives $\cos 	heta = \frac{1}{2}$ or $\cos 	heta = 2$.Since the range of $\cos 	heta$ is $[-1, 1]$,we reject $\cos 	heta = 2$.Thus,$\cos 	heta = \frac{1}{2} = \cos \left( \frac{\pi}{3} ight)$.The general solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm \alpha$.Therefore,$	heta = 2n\pi \pm \frac{\pi}{3}$.

If $\cos \theta + \sec \theta = \frac{5}{2}$,then the general value of $\theta$ is

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