Question:
If $\cos 4 x=1+k \sin ^{2} x \cos ^{2} x$, then write the value of $k$.
Solution:
We have,
$\cos 4 x=1+k \sin ^{2} \mathrm{x} \cos ^{2} x$
$\Rightarrow \cos (2 \times 2 x)=1+k \sin ^{2} x \cos ^{2} x$
$\Rightarrow 1-2 \sin ^{2} 2 x=1+k \sin ^{2} x \cos ^{2} x$
$\Rightarrow 1-2(2 \sin x \cos x)^{2}=1+k \sin ^{2} x \cos ^{2} x$
$\Rightarrow 1-8 \sin ^{2} x \cos ^{2} x=1+k \sin ^{2} x \cos ^{2} x$
$\Rightarrow \sin ^{2} x \cos ^{2} x(k+8)=0$
$\Rightarrow k+8=0$
$\therefore k=-8$