Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

Answer: (C)

$\frac{3}{4}$

${\cos }^6\alpha +{\sin }^6\alpha +k{\sin }^{2}2\alpha =1$

$\Rightarrow ({\cos }^2\alpha {)}^3+({\sin }^2\alpha {)}^3+k{\sin }^{2}2\alpha =1$

$\Rightarrow 1-3{\sin }^2\alpha {\cos }^2\alpha +k{\sin }^{2}2\alpha =1$

$\Rightarrow 1-\frac{3}{4}{\sin }^{2}2\alpha +k{\sin }^{2}2\alpha =1$

$\Rightarrow 1+{\sin }^{2}2\alpha (k-\frac{3}{4})=1$

$\Rightarrow k=\frac{3}{4}$