A) \[\frac{(1-\sqrt{7})}{4}\]
B) \[\frac{(4-\sqrt{7})}{3}\]
C) \[-\frac{(4+\sqrt{7})}{3}\]
D) \[\frac{(1+\sqrt{7})}{4}\]
Correct Answer: C
Solution :
Given \[\cos x+\sin x=\frac{1}{2}\Rightarrow 1+\sin 2x=\frac{1}{4}\] \[\Rightarrow \,\,\sin 2x=-\frac{3}{4},\] so x is obtuse and \[\frac{2\tan x}{1+{{\tan }^{2}}x}=\frac{3}{4}\] \[\Rightarrow \,\,3{{\tan }^{2}}x+8\tan x+3=0\] \[\therefore \,\,\tan x=\frac{-8\pm \sqrt{64-36}}{6}=-\frac{-4\pm \sqrt{7}}{3}\] as \[\tan x<0\]\[\therefore \,\,\tan x=\frac{-4-\sqrt{7}}{3}\]
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