If $3 \sin ^{2} x-8 \sin x+4=0, x \in\left(\frac{\pi}{2}, \pi ight)$, then $ an x=$

Question

If $3 \sin ^{2} x-8 \sin x+4=0, x \in\left(\frac{\pi}{2}, \piight)$, then $	an x=$

Aicharya · Accepted Answer

$3 \sin ^{2} x-8 \sin x+4=0$ $\sin x-2)\left(\sin x-\frac{2}{3}ight)=0$ $\sin x=2$ $x$ $\sin x=\frac{2}{3}$ $	an x=\frac{2}{\sqrt{5}}$ when $x \in(\pi / 2, \pi)$

If $3 \sin ^{2} x-8 \sin x+4=0, x \in\left(\frac{\pi}{2}, \pi\right)$, then $\tan x=$

Solution