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

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Question

(b) $2{\cos ^2}	heta - (\sqrt 2 + 1)\cos 	heta - 1 + \frac{{(\sqrt 2 + 1)}}{{\sqrt 2 }} = 0$

$ \Rightarrow $ $\cos 	heta = \frac{{(\sqrt 2 + 1)

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $2{\cos ^2}	heta - (\sqrt 2 + 1)\cos 	heta - 1 + \frac{{(\sqrt 2 + 1)}}{{\sqrt 2 }} = 0$

$ \Rightarrow $ $\cos 	heta = \frac{{(\sqrt 2 + 1) \pm \sqrt {{{(\sqrt 2 + 1)}^2} - \frac{8}{{\sqrt 2 }}} }}{4}$

$ \Rightarrow $ $\cos 	heta = \cos \left( {\frac{\pi }{4}} ight)$ 

$ \Rightarrow $ $	heta = 2n\pi \pm \frac{\pi }{4}$.

Trick : Since $	heta = \frac{\pi }{4}$ satisfies the equation and therefore the general value should be $2n\pi \pm \frac{\pi }{4}$.

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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