Question Number 179449 by Shrinava last updated on 29/Oct/22 | ||
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{3sinx}\:+\:\mathrm{4}\left(\mathrm{y}\:+\:\mathrm{cosx}\right)\:=\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{9} \\ $$ | ||
Answered by mr W last updated on 29/Oct/22 | ||
$$\mathrm{3}\:\mathrm{sin}\:{x}+\mathrm{4}\:\mathrm{cos}\:{x}−\mathrm{5}={y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{4} \\ $$$$\mathrm{5}\left[\mathrm{sin}\:\left({x}+\alpha\right)−\mathrm{1}\right]=\left({y}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${since}\:\mathrm{sin}\:\left({x}+\alpha\right)\leqslant\mathrm{1},\:{LHS}\:\leqslant\mathrm{0}. \\ $$$${but}\:{RHS}\geqslant\mathrm{0},\:{therefore} \\ $$$${LHS}={RHS}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\left({x}+\alpha\right)=\mathrm{1}\:\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow{y}=\mathrm{2} \\ $$ | ||
Commented by Shrinava last updated on 01/Nov/22 | ||
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$ | ||
Answered by manxsol last updated on 29/Oct/22 | ||