If cos^6 alpha + sin^6 alpha + K sin^2 2alpha | vedclass

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(B) Given the equation: $\cos^6 \alpha + \sin^6 \alpha + K \sin^2 2\alpha = 1$.We know that $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \cos^2 \alph

Vedclass · Accepted Answer

$\frac{3}{4}$

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(B) Given the equation: $\cos^6 \alpha + \sin^6 \alpha + K \sin^2 2\alpha = 1$.We know that $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \cos^2 \alpha$ and $b = \sin^2 \alpha$. Then $a + b = \cos^2 \alpha + \sin^2 \alpha = 1$.So,$\cos^6 \alpha + \sin^6 \alpha = (\cos^2 \alpha)^3 + (\sin^2 \alpha)^3 = (\cos^2 \alpha + \sin^2 \alpha)^3 - 3 \cos^2 \alpha \sin^2 \alpha (\cos^2 \alpha + \sin^2 \alpha)$.$= 1^3 - 3 \cos^2 \alpha \sin^2 \alpha (1) = 1 - 3 \cos^2 \alpha \sin^2 \alpha$.Substitute this into the original equation:$1 - 3 \cos^2 \alpha \sin^2 \alpha + K \sin^2 2\alpha = 1$.$-3 \cos^2 \alpha \sin^2 \alpha + K (2 \sin \alpha \cos \alpha)^2 = 0$.$-3 \cos^2 \alpha \sin^2 \alpha + K (4 \sin^2 \alpha \cos^2 \alpha) = 0$.Divide by $\sin^2 \alpha \cos^2 \alpha$ (assuming $\sin \alpha \cos \alpha 
eq 0$):$-3 + 4K = 0$.$4K = 3$,which gives $K = \frac{3}{4}$.

If $\cos^6 \alpha + \sin^6 \alpha + K \sin^2 2\alpha = 1,$ then $K =$

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