Question-82456

Question Number 82456 by liki last updated on 21/Feb/20

Commented by abdomathmax last updated on 21/Feb/20

cos(5x)=sin(4x)⇔cos(5x)=cos((π/2)−4x) ⇒ 5x=(π/2)−4x +2kπ or 5x=−(π/2) +4x +2kπ ⇒ 9x =(π/2) +2kπ or x=−(π/2) +2kπ (k∈Z) ⇒ x=(π/(18)) +((2kπ)/9) or x=−(π/2) +2kπ S ={(π/(18)) +((2kπ)/9) ,kεZ}∪{(2k−(1/2))π ,k∈Z}

$${cos}\left(\mathrm{5}{x}\right)={sin}\left(\mathrm{4}{x}\right)\Leftrightarrow{cos}\left(\mathrm{5}{x}\right)={cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\right)\:\Rightarrow \\ $$$$\mathrm{5}{x}=\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\:+\mathrm{2}{k}\pi\:{or}\:\mathrm{5}{x}=−\frac{\pi}{\mathrm{2}}\:+\mathrm{4}{x}\:+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$$\mathrm{9}{x}\:=\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:{or}\:{x}=−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:\:\left({k}\in{Z}\right)\:\Rightarrow \\ $$$${x}=\frac{\pi}{\mathrm{18}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{9}}\:{or}\:{x}=−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi \\ $$$${S}\:=\left\{\frac{\pi}{\mathrm{18}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{9}}\:,{k}\epsilon{Z}\right\}\cup\left\{\left(\mathrm{2}{k}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi\:,{k}\in{Z}\right\} \\ $$

Commented by jagoll last updated on 21/Feb/20

cos 5x = sin ((π/2)−5x) ⇒sin ((π/2)−5x) = sin 4x (1) (π/2)−5x+2kπ = 4x 9x = (π/2)+2kπ ⇒ x = (π/(18))+((2kπ)/9) (2) (π/2)+5x = 4x+ 2kπ x = −(π/2)+2kπ

$$\mathrm{cos}\:\mathrm{5}{x}\:=\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{5}{x}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{5}{x}\right)\:=\:\mathrm{sin}\:\mathrm{4}{x} \\ $$$$\left(\mathrm{1}\right)\:\frac{\pi}{\mathrm{2}}−\mathrm{5}{x}+\mathrm{2}{k}\pi\:=\:\mathrm{4}{x} \\ $$$$\mathrm{9}{x}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:\Rightarrow\:{x}\:=\:\frac{\pi}{\mathrm{18}}+\frac{\mathrm{2}{k}\pi}{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\pi}{\mathrm{2}}+\mathrm{5}{x}\:=\:\mathrm{4}{x}+\:\mathrm{2}{k}\pi\: \\ $$$${x}\:=\:−\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi \\ $$

Commented by liki last updated on 21/Feb/20