If 2 sin^{2} x + 7 cos x = 5,then the permiss | vedclass

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(A) Given equation: $2 \sin^{2} x + 7 \cos x = 5$ Since $\sin^{2} x = 1 - \cos^{2} x$,we substitute this into the equation: $2(1 - \cos^{2} x) + 7 \co

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$\frac{1}{2}$

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(A) Given equation: $2 \sin^{2} x + 7 \cos x = 5$ Since $\sin^{2} x = 1 - \cos^{2} x$,we substitute this into the equation: $2(1 - \cos^{2} x) + 7 \cos x = 5$ $2 - 2 \cos^{2} x + 7 \cos x = 5$ $2 \cos^{2} x - 7 \cos x + 3 = 0$ Let $t = \cos x$. Then $2t^{2} - 7t + 3 = 0$ $2t^{2} - 6t - t + 3 = 0$ $2t(t - 3) - 1(t - 3) = 0$ $(2t - 1)(t - 3) = 0$ So,$t = \frac{1}{2}$ or $t = 3$. Since the range of $\cos x$ is $[-1, 1]$,$t = 3$ is not possible. Therefore,$\cos x = \frac{1}{2}$.

If $2 \sin^{2} x + 7 \cos x = 5$,then the permissible value of $\cos x$ is

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