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Question Number 161060 by pete last updated on 11/Dec/21

Given sin(5x−38)=cos(2x+16), 0°≤x≤90°, find the value of x

$$\mathrm{Given}\:\mathrm{sin}\left(\mathrm{5x}−\mathrm{38}\right)=\mathrm{cos}\left(\mathrm{2x}+\mathrm{16}\right),\:\mathrm{0}°\leqslant\mathrm{x}\leqslant\mathrm{90}°, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$

Commented by cortano last updated on 11/Dec/21

recall sin (90°−x)= cos x ⇔ sin (5x−38°)= sin (90°−(128°−5x))=cos (128°−5x) ⇔cos (128°−5x)= cos (2x+16°) ⇔2x+16°= ± (128°−5x)+2nπ ⇔ { ((2x+16°=128°−5x+2nπ)),((2x+16°=5x−128°+2nπ)) :} ⇔ { ((7x=112°+2nπ)),((3x=144°+2kπ)) :} ⇔ { ((x=16°+((2nπ)/7))),((x=48°+((2kπ)/3))) :}

$$\:{recall}\:\mathrm{sin}\:\left(\mathrm{90}°−{x}\right)=\:\mathrm{cos}\:{x} \\ $$$$\Leftrightarrow\:\mathrm{sin}\:\left(\mathrm{5}{x}−\mathrm{38}°\right)=\:\mathrm{sin}\:\left(\mathrm{90}°−\left(\mathrm{128}°−\mathrm{5}{x}\right)\right)=\mathrm{cos}\:\left(\mathrm{128}°−\mathrm{5}{x}\right) \\ $$$$\Leftrightarrow\mathrm{cos}\:\left(\mathrm{128}°−\mathrm{5}{x}\right)=\:\mathrm{cos}\:\left(\mathrm{2}{x}+\mathrm{16}°\right) \\ $$$$\Leftrightarrow\mathrm{2}{x}+\mathrm{16}°=\:\pm\:\left(\mathrm{128}°−\mathrm{5}{x}\right)+\mathrm{2}{n}\pi \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{2}{x}+\mathrm{16}°=\mathrm{128}°−\mathrm{5}{x}+\mathrm{2}{n}\pi}\\{\mathrm{2}{x}+\mathrm{16}°=\mathrm{5}{x}−\mathrm{128}°+\mathrm{2}{n}\pi}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{7}{x}=\mathrm{112}°+\mathrm{2}{n}\pi}\\{\mathrm{3}{x}=\mathrm{144}°+\mathrm{2}{k}\pi}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{{x}=\mathrm{16}°+\frac{\mathrm{2}{n}\pi}{\mathrm{7}}}\\{{x}=\mathrm{48}°+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}}\end{cases} \\ $$

Commented by pete last updated on 11/Dec/21

thanks very much sir

$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 11/Dec/21

sin (5x−38)=cos (2x+16) cos (90−5x+38)=cos (2x+16) 90−5x+38=360k±(2x+16) 90−5x+38=360k+2x+16 112=360k+7x ⇒x=((360k)/7)+16=16°, 67(3/7)° 90−5x+38=360k−2x−16 144=360k+3x ⇒x=120k+48=48°

$$\mathrm{sin}\:\left(\mathrm{5}{x}−\mathrm{38}\right)=\mathrm{cos}\:\left(\mathrm{2}{x}+\mathrm{16}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{90}−\mathrm{5}{x}+\mathrm{38}\right)=\mathrm{cos}\:\left(\mathrm{2}{x}+\mathrm{16}\right) \\ $$$$\mathrm{90}−\mathrm{5}{x}+\mathrm{38}=\mathrm{360}{k}\pm\left(\mathrm{2}{x}+\mathrm{16}\right) \\ $$$$ \\ $$$$\mathrm{90}−\mathrm{5}{x}+\mathrm{38}=\mathrm{360}{k}+\mathrm{2}{x}+\mathrm{16} \\ $$$$\mathrm{112}=\mathrm{360}{k}+\mathrm{7}{x} \\ $$$$\Rightarrow{x}=\frac{\mathrm{360}{k}}{\mathrm{7}}+\mathrm{16}=\mathrm{16}°,\:\mathrm{67}\frac{\mathrm{3}}{\mathrm{7}}° \\ $$$$\mathrm{90}−\mathrm{5}{x}+\mathrm{38}=\mathrm{360}{k}−\mathrm{2}{x}−\mathrm{16} \\ $$$$\mathrm{144}=\mathrm{360}{k}+\mathrm{3}{x} \\ $$$$\Rightarrow{x}=\mathrm{120}{k}+\mathrm{48}=\mathrm{48}° \\ $$

Commented by pete last updated on 11/Dec/21

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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