Explanation

(d)
Given, $\cos{2x = \sqrt{2}\cos{x - 1 + \cos{x - \frac{1}{\sqrt{2}}}}}$
$\Rightarrow \ \ 1 + \cos{2x = \cos{x\left( \sqrt{2} + 1 \right) - \frac{1}{\sqrt{2}}}}$
$\Rightarrow \ 2\cos^{2}{x - \cos{x\left( \sqrt{2} + 1 \right) + \frac{1}{\sqrt{2}} = 0}}$
$\Rightarrow \ \cos{x = \frac{(\sqrt{2} + 1) \pm \sqrt{\left( \sqrt{2} + 1 \right)^{2} - \frac{8}{\sqrt{2}}}}{2(2)}}$
$= \frac{(\sqrt{2} + 1) \pm \sqrt{3 + 2\sqrt{2} - 4\sqrt{2}}}{4}$
$= \frac{\sqrt{2} + 1 \pm (\sqrt{2} - 1)}{4}$
$\Rightarrow \ \cos{x = \frac{\sqrt{2} + 1 + \sqrt{2} - 1}{4} = \frac{1}{\sqrt{2}}\ \ \ \left\lbrack \because\cos{x \neq \frac{1}{2}} \right\rbrack}$
$\Rightarrow \ \ x = 2n\pi \pm \frac{\pi}{4},\ \forall\ n \in Z$