(b) We have $1 – \cos 2x + 1 – {\cos ^2}2x = 2$

or $\cos 2x(\cos 2x + 1) = 0$

$\therefore $ $\cos 2x = 0,\, – 1$, 

$\therefore $ $2x = \left( {n + \frac{1}{2}} \right){\rm{ }}\pi \,{\rm{\, or\,\,}}\,\,(2n + 1){\rm{ }}\pi $

$ \Rightarrow $ $x = (2n + 1)\frac{\pi }{4}{\rm{\,\, or\,\, }}(2n + 1)\frac{\pi }{2}$

Now put $n = – 2,\, – 1,\,0,\,1,\,2$

$\therefore $ $x = \frac{{ – 3\pi }}{4},\,\frac{{ – \pi }}{4},\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4}$ 

and $\frac{{ – 3\pi }}{2},\,\frac{{ – \pi }}{2},\,\frac{\pi }{2},\,\frac{{3\pi }}{2},\,\frac{{5\pi }}{2}$

Since $ – \pi \le x \le \pi $, 

therefore $x \pm \frac{\pi }{4},\, \pm \frac{\pi }{2},\, \pm \frac{{3\pi }}{4}$ only.