Question Number 117387 by otchereabdullai@gmail.com last updated on 11/Oct/20
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{equation} \\ $$$$\mathrm{5sin}\theta+\mathrm{3}=\mathrm{0}\:\mathrm{for}\:\mathrm{value}\:\mathrm{of}\:\theta\:\mathrm{from}\:\mathrm{0}°\:\mathrm{to} \\ $$$$\mathrm{360}° \\ $$
Commented by bemath last updated on 11/Oct/20
$$\theta\:=\:\mathrm{sin}^{−\mathrm{1}} \:\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{k}.\mathrm{360}° \\ $$$$\theta\:=\:\mathrm{180}°−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{k}.\mathrm{360}° \\ $$$$\mathrm{where}\:\mathrm{sin}^{−\mathrm{1}} \left(−\frac{\mathrm{3}}{\mathrm{5}}\right)=\:−\mathrm{0}.\mathrm{643501} \\ $$
Answered by TANMAY PANACEA last updated on 11/Oct/20
$${sin}\theta=\frac{−\mathrm{3}}{\mathrm{5}}=−\mathrm{0}.\mathrm{6}=−{sin}\mathrm{37}^{{o}} \\ $$$${sin}\theta={sin}\left(\mathrm{180}^{{o}} +\mathrm{37}^{{o}} \right)={sin}\mathrm{217}^{{o}} \\ $$$${sin}\theta={sin}\left(\mathrm{360}^{{o}} −\mathrm{37}^{{o}} \right)={sin}\mathrm{323}^{{o}} \\ $$