If cos 2theta = (sqrt{2} + 1) left( cos theta | vedclass

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(B) Given equation: $\cos 2	heta = (\sqrt{2} + 1) \left( \cos 	heta - \frac{1}{\sqrt{2}} ight)$Using $\cos 2	heta = 2\cos^2 	heta - 1$,we get:$2

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$2n\pi \pm \frac{\pi}{4}$

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(B) Given equation: $\cos 2	heta = (\sqrt{2} + 1) \left( \cos 	heta - \frac{1}{\sqrt{2}} ight)$Using $\cos 2	heta = 2\cos^2 	heta - 1$,we get:$2\cos^2 	heta - 1 = (\sqrt{2} + 1)\cos 	heta - \frac{\sqrt{2} + 1}{\sqrt{2}}$$2\cos^2 	heta - (\sqrt{2} + 1)\cos 	heta - 1 + \frac{\sqrt{2} + 1}{\sqrt{2}} = 0$$2\cos^2 	heta - (\sqrt{2} + 1)\cos 	heta - 1 + 1 + \frac{1}{\sqrt{2}} = 0$$2\cos^2 	heta - (\sqrt{2} + 1)\cos 	heta + \frac{1}{\sqrt{2}} = 0$Multiplying by $\sqrt{2}$:$2\sqrt{2}\cos^2 	heta - (2 + \sqrt{2})\cos 	heta + 1 = 0$$(2\cos 	heta - 1)(\sqrt{2}\cos 	heta - 1) = 0$So,$\cos 	heta = \frac{1}{2}$ or $\cos 	heta = \frac{1}{\sqrt{2}}$.For $\cos 	heta = \frac{1}{\sqrt{2}}$,$	heta = 2n\pi \pm \frac{\pi}{4}$.Checking the options,the general solution is $2n\pi \pm \frac{\pi}{4}$.

If $\cos 2\theta = (\sqrt{2} + 1) \left( \cos \theta - \frac{1}{\sqrt{2}} \right)$,then the value of $\theta$ is

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