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If $3 \sin ^{2} x-8 \sin x+4=0, x \in\left(\frac{\pi}{2}, \pi\right)$, then $\tan x=$
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The correct answer is:
$-\frac{2}{\sqrt{5}}$
$3 \sin ^{2} x-8 \sin x+4=0$
$\sin x-2)\left(\sin x-\frac{2}{3}\right)=0$
$\sin x=2$
$x$
$\sin x=\frac{2}{3}$
$\tan x=\frac{2}{\sqrt{5}}$ when $x \in(\pi / 2, \pi)$
$\sin x-2)\left(\sin x-\frac{2}{3}\right)=0$
$\sin x=2$
$x$
$\sin x=\frac{2}{3}$
$\tan x=\frac{2}{\sqrt{5}}$ when $x \in(\pi / 2, \pi)$
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