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

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(D) 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

Vedclass · Accepted Answer

$\{2n\pi \pm \frac{\pi}{4} : n \in Z\}$

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(D) Given equation: $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$.
Using $\cos 2x = 2\cos^2 x - 1$,we have:
$2\cos^2 x - 1 = \sqrt{2}\cos x - 1 + \cos x - \frac{1}{\sqrt{2}}$
$2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$.
Using the quadratic formula $\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\cos x = \frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}+1)^2 - 4(2)(\frac{1}{\sqrt{2}})}}{4}$
$\cos x = \frac{(\sqrt{2}+1) \pm \sqrt{3+2\sqrt{2} - 4\sqrt{2}}}{4} = \frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}-1)^2}}{4}$
$\cos x = \frac{\sqrt{2}+1 \pm (\sqrt{2}-1)}{4}$.
Case $1$: $\cos x = \frac{\sqrt{2}+1 + \sqrt{2}-1}{4} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$.
However,the problem states $\cos x 
eq \frac{1}{\sqrt{2}}$,so this case is rejected.
Case $2$: $\cos x = \frac{\sqrt{2}+1 - (\sqrt{2}-1)}{4} = \frac{2}{4} = \frac{1}{2}$.
However,the problem states $\cos x 
eq \frac{1}{2}$.
Wait,re-evaluating the quadratic: $2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$.
Multiplying by $\sqrt{2}$: $2\sqrt{2}\cos^2 x - (2+\sqrt{2})\cos x + 1 = 0$.
$(2\cos x - 1)(\sqrt{2}\cos x - 1) = 0$.
Thus $\cos x = \frac{1}{2}$ or $\cos x = \frac{1}{\sqrt{2}}$.
Since both are excluded by the problem statement,there is no solution for $x$.

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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