Question

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

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

Answer 1

The correct option is A $\frac{\left(-4\pm \sqrt{7}\right)}{3}$

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

Divide both sides by $\cos x$

$1+\tan x=\frac{1}{2\cos x}$

Square both sides

$(1+\tan x{)}^2=\frac{{\sec }^{2}x}{4}$

$(1+{\tan }^{2}x+2\tan x)4={\sec }^{2}x=1+{\tan }^{2}x$

After simplification

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

$D=d^2=b^2-4ac=64-36=28$

$d=\pm 2\sqrt{7}$

These are $2$ real roots

$\tan x=\frac{-b}{2a}\pm \frac{d}{2a}=\frac{-8}{6}\pm \frac{2\sqrt{7}}{6}$

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