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Question Number 134538 by bemath last updated on 05/Mar/21

$$ \\ $$ How do you find the interval in which x belongs such that |sinx+cosx|=sinx + cosx?\\n

Answered by EDWIN88 last updated on 05/Mar/21

let sin x+cos x = y ⇒we get equation ∣y∣ = y this equation satisfied if only if y≥0 since sin x+cos x = (√2) sin (x+(π/4)) we get ⇒(√2) sin (x+(π/4)) ≥ 0 , sin (x+(π/4))≥ 0 note that sine function be ≥ 0 in first and 2^(nd) quadrant so 0 ≤ x+(π/4)≤π ; or −(π/4)+2kπ≤x≤((3π)/4) +2kπ.

$$\:\mathrm{let}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{y}\:\Rightarrow\mathrm{we}\:\mathrm{get}\:\mathrm{equation}\:\mid\mathrm{y}\mid\:=\:\mathrm{y}\: \\ $$ $$\mathrm{this}\:\mathrm{equation}\:\mathrm{satisfied}\:\mathrm{if}\:\mathrm{only}\:\mathrm{if}\:\mathrm{y}\geqslant\mathrm{0} \\ $$ $$\mathrm{\color{mathblue}{s}\color{mathblue}{i}\color{mathblue}{n}\color{mathblue}{c}\color{mathblue}{e}}\color{mathblue}{\:}\mathrm{\color{mathblue}{s}\color{mathblue}{i}\color{mathblue}{n}}\color{mathblue}{\:}\mathrm{\color{mathblue}{x}}\color{mathblue}{+}\mathrm{\color{mathblue}{c}\color{mathblue}{o}\color{mathblue}{s}}\color{mathblue}{\:}\mathrm{\color{mathblue}{x}}\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}\sqrt{\mathrm{\color{mathblue}{2}}}\color{mathblue}{\:}\mathrm{\color{mathblue}{s}\color{mathblue}{i}\color{mathblue}{n}}\color{mathblue}{\:}\color{mathblue}{\left(}\mathrm{\color{mathblue}{x}}\color{mathblue}{+}\frac{\color{mathblue}{\pi}}{\mathrm{\color{mathblue}{4}}}\color{mathblue}{\right)}\color{mathblue}{\:}\mathrm{\color{mathblue}{w}\color{mathblue}{e}}\color{mathblue}{\:}\mathrm{\color{mathblue}{g}\color{mathblue}{e}\color{mathblue}{t}} \\ $$ $$\color{mathblue}{\Rightarrow}\sqrt{\mathrm{\color{mathred}{2}}}\color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\color{mathred}{\left(}\mathrm{\color{mathred}{x}}\color{mathred}{+}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{4}}}\color{mathred}{\right)}\color{mathred}{\:}\color{mathred}{\geqslant}\color{mathred}{\:}\mathrm{\color{mathred}{0}}\color{mathred}{\:}\color{mathred}{,}\color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\color{mathred}{\left(}\mathrm{\color{mathred}{x}}\color{mathred}{+}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{4}}}\color{mathred}{\right)}\color{mathred}{\geqslant}\color{mathred}{\:}\mathrm{\color{mathred}{0}} \\ $$ $$\mathrm{\color{mathred}{n}\color{mathred}{o}\color{mathred}{t}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{t}\color{mathred}{h}\color{mathred}{a}\color{mathred}{t}}\color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{f}\color{mathred}{u}\color{mathred}{n}\color{mathred}{c}\color{mathred}{t}\color{mathred}{i}\color{mathred}{o}\color{mathred}{n}}\color{mathred}{\:}\mathrm{\color{mathred}{b}\color{mathred}{e}}\color{mathred}{\:}\color{mathred}{\geqslant}\color{mathred}{\:}\mathrm{\color{mathred}{0}}\color{mathred}{\:}\mathrm{\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\mathrm{\color{mathred}{f}\color{mathred}{i}\color{mathred}{r}\color{mathred}{s}\color{mathred}{t}}\color{mathred}{\:}\mathrm{\color{mathred}{a}\color{mathred}{n}\color{mathred}{d}}\color{mathred}{\:} \\ $$ $$\mathrm{\color{mathred}{2}}^{\mathrm{\color{mathred}{n}\color{mathred}{d}}} \color{mathred}{\:}\mathrm{\color{mathred}{q}\color{mathred}{u}\color{mathred}{a}\color{mathred}{d}\color{mathred}{r}\color{mathred}{a}\color{mathred}{n}\color{mathred}{t}}\color{mathbrown}{\:}\mathrm{\color{mathred}{s}\color{mathred}{o}}\color{mathred}{\:}\mathrm{\color{mathbrown}{0}}\color{mathbrown}{\:}\color{mathbrown}{\leqslant}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{x}}\color{mathbrown}{+}\frac{\color{mathbrown}{\pi}}{\mathrm{\color{mathbrown}{4}}}\color{mathbrown}{\leqslant}\color{mathbrown}{\pi}\color{mathbrown}{\:}\color{mathbrown}{;}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{o}\color{mathbrown}{r}}\color{mathbrown}{\:} \\ $$ $$\color{mathbrown}{−}\frac{\color{mathbrown}{\pi}}{\mathrm{\color{mathbrown}{4}}}+\mathrm{2k}\pi\color{mathbrown}{\leqslant}\mathrm{\color{mathbrown}{x}}\color{mathbrown}{\leqslant}\frac{\mathrm{\color{mathbrown}{3}}\color{mathbrown}{\pi}}{\mathrm{\color{mathbrown}{4}}}\color{mathbrown}{\:}+\mathrm{2k}\pi\color{mathbrown}{.} \\ $$

Commented byEDWIN88 last updated on 05/Mar/21

Commented byEDWIN88 last updated on 05/Mar/21

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