Question

If ${\cos }^6\alpha +{\sin }^6\alpha +k{\sin }^{2}2\alpha =1\forall \alpha \in \left(0,\pi /2\right)$, then k is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The given condition can be written
${\left({\cos }^2\alpha +{\sin }^2\alpha \right)}^3-3{\sin }^2\alpha {\cos }^2\alpha \left({\cos }^2\alpha +{\sin }^2\alpha \right)+k{\sin }^{2}2\alpha =1$
$\Rightarrow \left(-\frac{3}{4}\right){\sin }^{2}2\alpha +k{\sin }^{2}2\alpha =0,$
Showing that $k=\frac{3}{4}$.