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Question Number 207588 by hardmath last updated on 19/May/24

(√2) sinx + cosx ≥ 1

$$\sqrt{\mathrm{2}}\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{cos}\boldsymbol{\mathrm{x}}\:\:\geqslant\:\:\mathrm{1} \\ $$

Answered by mr W last updated on 19/May/24

((√2)/( (√3))) sin x+(1/( (√3))) cos x≥(1/( (√3))) sin α sin x+cos α cos x≥(1/( (√3))) with α=cos^(−1) (1/( (√3))) cos (x−α)≥(1/( (√3))) 2kπ−cos^(−1) (1/( (√3)))≤x−α≤2kπ+cos^(−1) (1/( (√3))) ⇒2kπ≤x≤2kπ+2 cos^(−1) (1/( (√3))) or ⇒2kπ≤x≤(2k+1)π−cos^(−1) (1/( 3))

$$\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}\:{x}\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{sin}\:\alpha\:\:\mathrm{sin}\:{x}+\mathrm{cos}\:\alpha\:\mathrm{cos}\:{x}\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${with}\:\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cos}\:\left({x}−\alpha\right)\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2}{k}\pi−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\leqslant{x}−\alpha\leqslant\mathrm{2}{k}\pi+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{2}{k}\pi\leqslant{x}\leqslant\mathrm{2}{k}\pi+\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{2}{k}\pi\leqslant{x}\leqslant\left(\mathrm{2}{k}+\mathrm{1}\right)\pi−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\mathrm{3}} \\ $$

Commented by hardmath last updated on 19/May/24

thank you very mych professor

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{mych}\:\mathrm{professor} \\ $$

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