If cos 2x = (sqrt{2}+1)(cos x - frac{1}{sqrt{ | vedclass

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(A) Given equation: $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$Using $\cos 2x = 2\cos^2 x - 1$, we have:$2\cos^2 x - 1 = \sqrt{2}\cos x - 1

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$\{2n\pi \pm \frac{\pi}{3} : n \in Z\}$

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(A) Given equation: $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$Using $\cos 2x = 2\cos^2 x - 1$, we have:$2\cos^2 x - 1 = \sqrt{2}\cos x - 1 + \cos x - \frac{1}{\sqrt{2}}$$2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$Multiplying by $\sqrt{2}$:$2\sqrt{2}\cos^2 x - (2+\sqrt{2})\cos x + 1 = 0$$2\sqrt{2}\cos^2 x - 2\cos x - \sqrt{2}\cos x + 1 = 0$$2\cos x(\sqrt{2}\cos x - 1) - 1(\sqrt{2}\cos x - 1) = 0$$(2\cos x - 1)(\sqrt{2}\cos x - 1) = 0$Thus, $\cos x = \frac{1}{2}$ or $\cos x = \frac{1}{\sqrt{2}}$.Since it is given that $\cos x 
eq \frac{1}{\sqrt{2}}$, we must have $\cos x = \frac{1}{2}$.$\cos x = \cos\left(\frac{\pi}{3}ight)$$\Rightarrow x = 2n\pi \pm \frac{\pi}{3}$, where $n \in Z$.

If $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$ and $\cos x \neq \frac{1}{\sqrt{2}}$,then $x \in$

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