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

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

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

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(A) Given equation: $\cos 2x = (\sqrt{2} + 1) \left(\cos x - \frac{1}{\sqrt{2}}ight)$.Using $\cos 2x = 2\cos^2 x - 1$,we get:$2\cos^2 x - 1 = (\sqrt{2} + 1)\cos x - \frac{\sqrt{2} + 1}{\sqrt{2}}$$2\cos^2 x - (\sqrt{2} + 1)\cos x - 1 + \frac{\sqrt{2} + 1}{\sqrt{2}} = 0$$2\cos^2 x - (\sqrt{2} + 1)\cos x - 1 + 1 + \frac{1}{\sqrt{2}} = 0$$2\cos^2 x - (\sqrt{2} + 1)\cos x + \frac{1}{\sqrt{2}} = 0$Using the quadratic formula $\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$\cos x = \frac{(\sqrt{2} + 1) \pm \sqrt{(\sqrt{2} + 1)^2 - 4(2)(\frac{1}{\sqrt{2}})}}{4}$$\cos x = \frac{(\sqrt{2} + 1) \pm \sqrt{3 + 2\sqrt{2} - 4\sqrt{2}}}{4} = \frac{(\sqrt{2} + 1) \pm \sqrt{3 - 2\sqrt{2}}}{4}$Since $3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$,we have:$\cos x = \frac{(\sqrt{2} + 1) \pm (\sqrt{2} - 1)}{4}$Case $1$: $\cos x = \frac{\sqrt{2} + 1 + \sqrt{2} - 1}{4} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$. (Rejected as $\cos x 
eq \frac{1}{\sqrt{2}}$)Case $2$: $\cos x = \frac{\sqrt{2} + 1 - \sqrt{2} + 1}{4} = \frac{2}{4} = \frac{1}{2}$.For $\cos x = \frac{1}{2}$,$x = 2n\pi \pm \frac{\pi}{3}$.

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

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