Question

If ${\cos }^6\alpha +{\sin }^6\alpha +K{\sin }^2\left(2\alpha \right)=1$, then $K=?$

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

Answer 1

The correct option is B

$\frac{3}{4}$

Find the value of $K$:

Given, ${\cos }^6\alpha +{\sin }^6\alpha +K{\sin }^2\left(2\alpha \right)=1$

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

$\Rightarrow$ $\left({\cos }^2\alpha +{\sin }^2\alpha \right)\left[{\left({\cos }^2\alpha \right)}^2+{\left({\sin }^2\alpha \right)}^2-{\cos }^2\alpha {\sin }^2\alpha \right]+K{\sin }^2\left(2\alpha \right)=1$

$\Rightarrow$$1\left[{\left({\cos }^2\alpha \right)}^2+{\left({\sin }^2\alpha \right)}^2+2{\sin }^2\alpha {\cos }^2\alpha -3{\sin }^2\alpha {\cos }^2\alpha \right]+K{\sin }^2\left(2\alpha \right)=1$

$\Rightarrow$ $1-3{\sin }^2\alpha {\cos }^2\alpha +K{\sin }^2\left(2\alpha \right)=1$

$\Rightarrow$ $1-\left(\frac{3}{4}\right){\left(2\sin \alpha \cos \alpha \right)}^2+K{\sin }^2\left(2\alpha \right)=1$

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

$\Rightarrow$ $\left(K-\frac{3}{4}\right)=0$

$\mathbf{\therefore }\mathit{K}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{4}}$

Hence, Option ‘B’ is Correct.