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Question Number 17148 by Tinkutara last updated on 01/Jul/17

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$\cos \left(\pi \sqrt{x-4}\right)\cos \left(\pi \sqrt{x}\right)=1\mathrm{is}$$

Answered by ajfour last updated on 01/Jul/17

$$\Rightarrow \cos \left(\pi \sqrt{\mathrm{x}-4}\right)=\pm 1$$$$\mathrm{and}\cos \left(\pi \sqrt{\mathrm{x}}\right)=\pm 1$$$$\mathrm{which}\mathrm{means}\mathrm{for}\mathrm{n},\mathrm{k}\in +\mathrm{ve}\mathrm{integers}$$$$\pi \sqrt{\mathrm{x}-4}=\mathrm{n}\pi \mathrm{and}\pi \sqrt{\mathrm{x}}=\mathrm{k}\pi$$$$\mathrm{both}\mathrm{n},\mathrm{k}\mathrm{simultaneously}$$$$\mathrm{odd}\mathrm{or}\mathrm{even}\left(\mathrm{including}\mathrm{zero}\right);$$$$\Rightarrow \mathrm{x}-4={\mathrm{n}}^2\mathrm{and}\mathrm{x}={\mathrm{k}}^2$$$$\mathrm{combining}\mathrm{them}\mathrm{we}\mathrm{have}:$$$${\mathrm{n}}^2+4={\mathrm{k}}^2$$$$\Rightarrow \mathrm{n}=0,\mathrm{k}=2$$$$\mathrm{and}\mathrm{none}\mathrm{other}..$$$$\mathrm{x}={\mathrm{k}}^2=4$$$$\mathrm{hence}\mathrm{x}=4\mathrm{seems}\mathrm{to}\mathrm{be}\mathrm{the}$$$$\mathrm{only}\mathrm{solution}.$$

Commented by Tinkutara last updated on 01/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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