Question Number 29491 by math solver last updated on 09/Feb/18
$${If}\:\mathrm{4}{sinx}.{cosy}+\mathrm{2}{sinx}+\mathrm{2}{cosy}+\mathrm{1}=\mathrm{0} \\ $$$${where}\:{x},{y}\:\in\:\left[\mathrm{0},\mathrm{2}{pie}\right].\:{Find}\:{largest}\: \\ $$$${possible}\:{value}\:{of}\:{the}\:{sum}\:\left({x}+{y}\right). \\ $$
Answered by ajfour last updated on 09/Feb/18
$$\left(\mathrm{1}+\mathrm{2sin}\:{x}\right)\left(\mathrm{1}+\mathrm{2cos}\:{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}\:;\:\:\mathrm{cos}\:{y}\:=\:{whatever} \\ $$$${or}\:\:\mathrm{cos}\:{y}=−\frac{\mathrm{1}}{\mathrm{2}};\:\:\mathrm{sin}\:{x}=\:{whatever} \\ $$$${in}\:{this}\:{case} \\ $$$$\left({x}+{y}\right)_{{max}} \:=\mathrm{2}\pi+\left(\pi+\frac{\pi}{\mathrm{3}}\right)\:=\frac{\mathrm{10}\pi}{\mathrm{3}} \\ $$$${under}\:{first}\:{case} \\ $$$$\left({x}+{y}\right)_{{max}} \:=\:\left(\mathrm{2}\pi−\frac{\pi}{\mathrm{6}}\right)+\mathrm{2}\pi\:=\:\frac{\mathrm{23}\pi}{\mathrm{6}} \\ $$$${Hence}\:\:\left({x}+{y}\right)_{{max}} \:=\frac{\mathrm{23}\pi}{\mathrm{6}}\:. \\ $$
Commented by math solver last updated on 09/Feb/18
$${thank}\:{you}\:{sir}. \\ $$