If cos

$\cos ^{6} x+\sin ^{6} x+k \sin ^{2} 2 x=1 \quad$ (Given)

L . H . S .

$=\left(\cos ^{2} x\right)^{3}+\left(\sin ^{2} x\right)^{3}+k \sin ^{2} 2 x$

$=\left(\cos ^{2} x+\sin ^{2} x\right)^{3}+k \sin ^{2} 2 x+3 \sin ^{2} x \cos ^{4} x+3 \sin ^{4} x \cos ^{2} x$

$\left[\right.$ Since $\left.(a+b)^{3}=a^{3}+b^{3}-3 a b(a+b)\right]$

$=(1)^{3}+3 \sin ^{2} x \cos ^{4} x+3 \sin ^{4} x \cos ^{2} x+k \sin ^{2} 2 x$

$=1+3 \sin ^{2} x \cos ^{2} x\left(\cos ^{2} x+\sin ^{2} x\right)+k \sin ^{2} 2 x$

$=1+3 \sin ^{2} x \cos ^{2} x+k \sin ^{2} 2 x$

$=1+\frac{3}{4}(2 \sin x \cos x)^{2}+k \sin ^{2} 2 x$

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

i. e. $1+\sin ^{2} 2 x\left(k+\frac{3}{4}\right)$

and R.H.S. is given 1

$=1+\sin ^{2} 2 \mathrm{x}\left(\mathrm{k}+\frac{3}{4}\right)=0$

$=\mathrm{k}=-3 / 4 \quad$ or $\quad \sin ^{2} 2 \mathrm{x}=0$