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Question Number 164328 by mnjuly1970 last updated on 16/Jan/22

$$$$$$solve$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{cos}^{2n}\left(x\right)}{n!}=\sqrt[4]{e}◼$$$$-------$$$$$$

Answered by qaz last updated on 16/Jan/22

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{\cos {}^{2\mathrm{n}}\mathrm{x}}{\mathrm{n}!}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}!}{\left(\frac{1+\cos 2\mathrm{x}}{2}\right)}^{\mathrm{n}}={\mathrm{e}}^{\frac{1+\cos 2\mathrm{x}}{2}}={\mathrm{e}}^{\frac{1}{4}}$$$$\Rightarrow \cos 2\mathrm{x}=-\frac{1}{2}$$$$\Rightarrow 2\mathrm{x}=(2\mathrm{k}+1)\pi \pm \frac{\pi }{3}$$$$\Rightarrow \mathrm{x}=\frac{(2\mathrm{k}+1)\pi }{2}\pm \frac{\pi }{6}...(\mathrm{k}\in \mathrm{Z})$$

Answered by Lordose last updated on 16/Jan/22

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}{\mathrm{n}!}=\sqrt[4]{\mathrm{e}}$$$${\mathrm{e}}^{{\cos }^2\left(\mathrm{x}\right)}={\mathrm{e}}^{\frac{1}{4}}$$$${\cos }^2\left(\mathrm{x}\right)=\frac{1}{4}$$$$\cos \left(\mathrm{x}\right)=\pm \frac{1}{2}$$$$\mathrm{x}={\cos }^{-1}(\pm \frac{1}{2})$$

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