If 3 sin^4 x + 2 cos^4 x = frac{6}{5} and x i | vedclass

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(B) Given $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.Substitute $\cos^2 x = 1 - \sin^2 x$:$3 \sin^4 x + 2(1 - \sin^2 x)^2 = \frac{6}{5}$$3 \sin^4 x + 2(1

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$2 \sqrt{6}$

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(B) Given $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.Substitute $\cos^2 x = 1 - \sin^2 x$:$3 \sin^4 x + 2(1 - \sin^2 x)^2 = \frac{6}{5}$$3 \sin^4 x + 2(1 + \sin^4 x - 2 \sin^2 x) = \frac{6}{5}$$5 \sin^4 x - 4 \sin^2 x + 2 = \frac{6}{5}$$25 \sin^4 x - 20 \sin^2 x + 10 = 6$$25 \sin^4 x - 20 \sin^2 x + 4 = 0$$(5 \sin^2 x - 2)^2 = 0$$\sin^2 x = \frac{2}{5}$,so $\cos^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.Now,$\sin 2x = 2 \sin x \cos x$,so $\sin^2 2x = 4 \sin^2 x \cos^2 x = 4 	imes \frac{2}{5} 	imes \frac{3}{5} = \frac{24}{25}$.$\sin 2x = \frac{2 \sqrt{6}}{5}$ (since $x$ is acute,$2x$ is in the first or second quadrant).$\cos 2x = 2 \cos^2 x - 1 = 2(\frac{3}{5}) - 1 = \frac{1}{5}$.$	an 2x = \frac{\sin 2x}{\cos 2x} = \frac{2 \sqrt{6} / 5}{1 / 5} = 2 \sqrt{6}$.

If $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$ and $x$ is an acute angle,then $\tan 2x =$

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