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

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Question

(b) Since ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$

using formula ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ and on solving,

w

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(b) Since ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$

using formula ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ and on solving,

we get the required result $i.e.$ $K = \frac{3}{4}$.

જો ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ તો $K =$

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