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

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(B) We are given the equation: ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$. Using the identity ${a^3} + {b^3} = {(a + b)^3} - 3ab(a +

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$\frac{3}{4}$

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(B) We are given the equation: ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$. Using the identity ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$,let $a = {\cos ^2}\alpha$ and $b = {\sin ^2}\alpha$. Then,${{\cos ^6}\alpha + {\sin ^6}\alpha = {(\cos ^2}\alpha + {\sin ^2}\alpha)^3 - 3{\cos ^2}\alpha {\sin ^2}\alpha (\cos ^2}\alpha + {\sin ^2}\alpha)$. Since ${\cos ^2}\alpha + {\sin ^2}\alpha = 1$,this simplifies to $1 - 3{\cos ^2}\alpha {\sin ^2}\alpha$. We know that $2\sin \alpha \cos \alpha = \sin 2\alpha$,so ${\sin ^2}2\alpha = 4{\sin ^2}\alpha {\cos ^2}\alpha$,which implies ${\sin ^2}\alpha {\cos ^2}\alpha = \frac{1}{4}{\sin ^2}2\alpha$. Substituting this back into the equation: $1 - 3(\frac{1}{4}{\sin ^2}2\alpha) + K{\sin ^2}2\alpha = 1$. $K{\sin ^2}2\alpha - \frac{3}{4}{\sin ^2}2\alpha = 0$. $(K - \frac{3}{4}){\sin ^2}2\alpha = 0$. For this to hold for all $\alpha$,we must have $K = \frac{3}{4}$.

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

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