यदि 3 sin^4 x + 2 cos^4 x = frac{6}{5} और x ए | vedclass

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(B) दिया गया है $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.$\cos^2 x = 1 - \sin^2 x$ प्रतिस्थापित करने पर:$3 \sin^4 x + 2(1 - \sin^2 x)^2 = \frac{6}{5}$$

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$2 \sqrt{6}$

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(B) दिया गया है $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.$\cos^2 x = 1 - \sin^2 x$ प्रतिस्थापित करने पर:$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}$$25 \sin^4 x - 20 \sin^2 x + 4 = 0$$(5 \sin^2 x - 2)^2 = 0$$\sin^2 x = \frac{2}{5}$,अतः $\cos^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.अब,$\sin 2x = 2 \sin x \cos x$,इसलिए $\sin^2 2x = 4 \sin^2 x \cos^2 x = 4 	imes \frac{2}{5} 	imes \frac{3}{5} = \frac{24}{25}$.$\sin 2x = \frac{2 \sqrt{6}}{5}$.$\cos 2x = 2 \cos^2 x - 1 = 2(\frac{3}{5}) - 1 = \frac{1}{5}$.$	an 2x = \frac{\sin 2x}{\cos 2x} = \frac{2 \sqrt{6} / 5}{1 / 5} = 2 \sqrt{6}$.

यदि $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$ और $x$ एक न्यून कोण है,तो $\tan 2x =$

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