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(C) Given $\cos x + \sin x = \frac{1}{2}$.Squaring both sides,we get $(\cos x + \sin x)^2 = \frac{1}{4}$.$1 + 2 \sin x \cos x = \frac{1}{4} \Rightarro

Vedclass · Accepted Answer

$-\frac{4 + \sqrt{7}}{3}$

Vedclass · Answer

(C) Given $\cos x + \sin x = \frac{1}{2}$.Squaring both sides,we get $(\cos x + \sin x)^2 = \frac{1}{4}$.$1 + 2 \sin x \cos x = \frac{1}{4} \Rightarrow 2 \sin x \cos x = -\frac{3}{4}$.Since $\cos x + \sin x = \frac{1}{2} > 0$ and $2 \sin x \cos x Using $\sin x + \cos x = \frac{1}{2}$,we have $\sin x - \cos x = \pm \sqrt{(\sin x + \cos x)^2 - 4 \sin x \cos x} = \pm \sqrt{\frac{1}{4} - 4(-\frac{3}{4})} = \pm \sqrt{\frac{1}{4} + 3} = \pm \sqrt{\frac{13}{4}} = \pm \frac{\sqrt{13}}{2}$.Since $x \in (\frac{\pi}{2}, \pi)$,$\sin x > 0$ and $\cos x  0$. Thus,$\sin x - \cos x = \frac{\sqrt{13}}{2}$.Adding the two equations: $2 \sin x = \frac{1 + \sqrt{13}}{2} \Rightarrow \sin x = \frac{1 + \sqrt{13}}{4}$.Subtracting: $2 \cos x = \frac{1 - \sqrt{13}}{2} \Rightarrow \cos x = \frac{1 - \sqrt{13}}{4}$.$	an x = \frac{\sin x}{\cos x} = \frac{1 + \sqrt{13}}{1 - \sqrt{13}} = \frac{(1 + \sqrt{13})^2}{1 - 13} = \frac{1 + 13 + 2\sqrt{13}}{-12} = \frac{14 + 2\sqrt{13}}{-12} = -\frac{7 + \sqrt{13}}{6}$.Wait,re-evaluating the quadratic equation approach: $3 	an^2 x + 8 	an x + 3 = 0$ leads to $	an x = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm 2\sqrt{7}}{6} = \frac{-4 \pm \sqrt{7}}{3}$.Since $x$ is in the second quadrant,$	an x < 0$. Both values are negative. Checking $\cos x + \sin x = \frac{1}{2}$ with $	an x = \frac{-4 - \sqrt{7}}{3}$ gives the correct result.

If $0 < x < \pi$ and $\cos x + \sin x = \frac{1}{2}$,then $\tan x$ is:

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