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Question Number 21050 by Tinkutara last updated on 10/Sep/17

$$\mathrm{The}\mathrm{most}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}$$$$\mathrm{equation}\sin x+\cos x=\underset{a\in R}{\min }\{1,a^2-4a+6\}$$$$\mathrm{is}$$

Answered by mrW1 last updated on 11/Sep/17

$${\mathrm{a}}^2-4\mathrm{a}+6={\mathrm{a}}^2-2\times 2\times \mathrm{a}+2^2+2={(\mathrm{a}-2)}^2+2⩾2>1$$$$\Rightarrow \underset{a\in R}{\min }\{1,a^2-4a+6\}=1$$$$\Rightarrow \sin \mathrm{x}+\cos \mathrm{x}=1$$$$\Rightarrow \sqrt{2}(\sin \mathrm{x}\cos \frac{\pi }{4}+\sin \frac{\pi }{4}\cos \mathrm{x})=1$$$$\Rightarrow \sqrt{2}\sin (\mathrm{x}+\frac{\pi }{4})=1$$$$\Rightarrow \sin (\mathrm{x}+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$$$$\Rightarrow \mathrm{x}+\frac{\pi }{4}=\mathrm{n}\pi +{(-1)}^{\mathrm{n}}\frac{\pi }{4}$$$$\Rightarrow \mathrm{x}=\mathrm{n}\pi +[{(-1)}^{\mathrm{n}}-1]\frac{\pi }{4}$$

Commented by Tinkutara last updated on 12/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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