If 3 sin^{2} x - 8 sin x + 4 = 0 and x in lef | vedclass

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Question

(C) Given the equation $3 \sin^{2} x - 8 \sin x + 4 = 0$. Let $u = \sin x$. Then $3u^{2} - 8u + 4 = 0$. Factoring the quadratic: $(3u - 2)(u - 2) = 0$

Vedclass · Accepted Answer

$-\frac{2}{\sqrt{5}}$

Vedclass · Answer

(C) Given the equation $3 \sin^{2} x - 8 \sin x + 4 = 0$. Let $u = \sin x$. Then $3u^{2} - 8u + 4 = 0$. Factoring the quadratic: $(3u - 2)(u - 2) = 0$. This gives $u = \frac{2}{3}$ or $u = 2$. Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is impossible. Thus,$\sin x = \frac{2}{3}$. Since $x \in \left(\frac{\pi}{2}, \piight)$,$x$ lies in the second quadrant where $\cos x$ is negative. Using $\cos^{2} x = 1 - \sin^{2} x = 1 - \left(\frac{2}{3}ight)^{2} = 1 - \frac{4}{9} = \frac{5}{9}$. Therefore,$\cos x = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$. Finally,$	an x = \frac{\sin x}{\cos x} = \frac{2/3}{-\sqrt{5}/3} = -\frac{2}{\sqrt{5}}$.

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

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