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

The most general solution of the equation sinx + cosx = min_(a∈R) {1, a^2 − 4a + 6} is

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

Answered by mrW1 last updated on 11/Sep/17

a^2 −4a+6=a^2 −2×2×a+2^2 +2=(a−2)^2 +2≥2>1 ⇒ min_(a∈R) {1, a^2 − 4a + 6}=1 ⇒sin x+cos x=1 ⇒(√2)(sin x cos (π/4)+sin (π/4) cos x)=1 ⇒(√2) sin (x+(π/4))=1 ⇒ sin (x+(π/4))=(1/(√2)) ⇒x+(π/4)=nπ+(−1)^n (π/4) ⇒x=nπ+[(−1)^n −1](π/4)

$$\mathrm{a}^{\mathrm{2}} −\mathrm{4a}+\mathrm{6}=\mathrm{a}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}×\mathrm{a}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}=\left(\mathrm{a}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}\geqslant\mathrm{2}>\mathrm{1} \\ $$$$\Rightarrow\:\underset{{a}\in{R}} {\mathrm{min}}\left\{\mathrm{1},\:{a}^{\mathrm{2}} \:−\:\mathrm{4}{a}\:+\:\mathrm{6}\right\}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}=\mathrm{1} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:\mathrm{x}\right)=\mathrm{1} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{x}+\frac{\pi}{\mathrm{4}}=\mathrm{n}\pi+\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{n}\pi+\left[\left(−\mathrm{1}\right)^{\mathrm{n}} −\mathrm{1}\right]\frac{\pi}{\mathrm{4}} \\ $$

Commented by Tinkutara last updated on 12/Sep/17

Thank you very much Sir!

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

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