Question

If $0<x<\pi ,$ and $\cos x+\sin x=1/2$, then $\tan x=$

• A. $(-4\pm \sqrt{7})/3$
• B. $(1+\sqrt{7})/4$
• C. $(1-\sqrt{7})/4$
• D. $(4-\sqrt{7})/3$

Answer 1

Answer: (A)

$(-4\pm \sqrt{7})/3$

$\cos x+\sin x=\frac{1}{2}\Rightarrow {(\cos x+\sin x)}^2=\frac{1}{4}$

${\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x=\frac{1}{4}\Rightarrow 1+\sin 2x=\frac{1}{4}$

$\sin 2x=-\frac{3}{4}\Rightarrow \frac{2\tan x}{1+{\tan }^{2}x}=-\frac{3}{4}$

$\Rightarrow 8\tan x=-3-3{\tan }^{2}x\Rightarrow 3{\tan }^{2}x+8\tan x+3=0$

$\tan x=\frac{-8\pm \sqrt{64-36}}{2\times 3}=\frac{-8\pm 2\sqrt{7}}{6}=\frac{-8\pm 2\sqrt{7}}{6}$

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