If $0 < x < \pi$ and $\cos x+\sin x=\frac{1}{2}$, then $\tan x$ is
If $0 < x < \pi$ and $\cos x+\sin x=\frac{1}{2}$, then $\tan x$ is
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$\frac{(1-\sqrt{7})}{4}$
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$\frac{(4-\sqrt{7})}{3}$
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$-\frac{(4+\sqrt{7})}{3}$
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$\frac{(1+\sqrt{7})}{4}$
Solution
$\cos x+\sin x=\frac{1}{2} \Rightarrow 1+\sin 2 x=\frac{1}{4} \Rightarrow \sin 2 x=-\frac{3}{4}$, so $x$ is obtuse and $\frac{2 \tan x}{1+\tan ^2 x}=-\frac{3}{4} \Rightarrow 3 \tan ^2 x+8 \tan x+3=0$
$\therefore \tan x=\frac{-8 \pm \sqrt{64-36}}{6}=\frac{-4 \pm \sqrt{7}}{3}$
$\because \tan x < 0 \quad \therefore \tan x=\frac{-4-\sqrt{7}}{3}$
Asked in: JEE Main 2006
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