(b) On squaring the given relation

$\sin 2\theta = {x^2} – 1 \le 1 \Rightarrow {x^2} \le 2$

or $ – \sqrt 2 \le x \le \sqrt 2 $          $[\because \,\,\sin 2\theta  \le 1]$

Now ${\sin ^6}\theta + {\cos ^6}\theta $

$ = {({\sin ^2}\theta + {\cos ^2}\theta )^3} – 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta + {\cos ^2}\theta )$

$ = 1 – 3{\sin ^2}\theta {\cos ^2}\theta = 1 – \frac{3}{4}{\sin ^2}2\theta $

$ = 1 – \frac{3}{4}{({x^2} – 1)^2} = \frac{1}{4}\{ 4 – 3{({x^2} – 1)^2}\} $

Thus the given result will hold true only when ${x^2} \le 2$ and not for all real values of $x.$