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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $\cos x + \sin x = \frac{1}{2}$.Squaring both sides,we get $(\cos x + \sin x)^2 = (\frac{1}{2})^2$.$\cos^2 x + \sin^2 x + 2 \sin x \cos x = \frac{1}{4}$.Since $\cos^2 x + \sin^2 x = 1$,we have $1 + \sin 2x = \frac{1}{4}$,which implies $\sin 2x = -\frac{3}{4}$.Since $\sin 2x Using the identity $\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$,we have $\frac{2 	an x}{1 + 	an^2 x} = -\frac{3}{4}$.$8 	an x = -3 - 3 	an^2 x \Rightarrow 3 	an^2 x + 8 	an x + 3 = 0$.Using the quadratic formula $	an x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $	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 $	an x For $	an x = \frac{-4 - \sqrt{7}}{3}$,$\cos x + \sin x$ is negative,which contradicts the given condition. Thus,$	an x = \frac{-4 + \sqrt{7}}{3}$.

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

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