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Question Number 29491 by math solver last updated on 09/Feb/18

If 4sinx.cosy+2sinx+2cosy+1=0 where x,y ∈ [0,2pie]. Find largest possible value of the sum (x+y).

$${If}\:\mathrm{4}{sinx}.{cosy}+\mathrm{2}{sinx}+\mathrm{2}{cosy}+\mathrm{1}=\mathrm{0} \\ $$$${where}\:{x},{y}\:\in\:\left[\mathrm{0},\mathrm{2}{pie}\right].\:{Find}\:{largest}\: \\ $$$${possible}\:{value}\:{of}\:{the}\:{sum}\:\left({x}+{y}\right). \\ $$

Answered by ajfour last updated on 09/Feb/18

(1+2sin x)(1+2cos y)=0 ⇒ sin x=−(1/2) ; cos y = whatever or cos y=−(1/2); sin x= whatever in this case (x+y)_(max) =2π+(π+(π/3)) =((10π)/3) under first case (x+y)_(max) = (2π−(π/6))+2π = ((23π)/6) Hence (x+y)_(max) =((23π)/6) .

$$\left(\mathrm{1}+\mathrm{2sin}\:{x}\right)\left(\mathrm{1}+\mathrm{2cos}\:{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}\:;\:\:\mathrm{cos}\:{y}\:=\:{whatever} \\ $$$${or}\:\:\mathrm{cos}\:{y}=−\frac{\mathrm{1}}{\mathrm{2}};\:\:\mathrm{sin}\:{x}=\:{whatever} \\ $$$${in}\:{this}\:{case} \\ $$$$\left({x}+{y}\right)_{{max}} \:=\mathrm{2}\pi+\left(\pi+\frac{\pi}{\mathrm{3}}\right)\:=\frac{\mathrm{10}\pi}{\mathrm{3}} \\ $$$${under}\:{first}\:{case} \\ $$$$\left({x}+{y}\right)_{{max}} \:=\:\left(\mathrm{2}\pi−\frac{\pi}{\mathrm{6}}\right)+\mathrm{2}\pi\:=\:\frac{\mathrm{23}\pi}{\mathrm{6}} \\ $$$${Hence}\:\:\left({x}+{y}\right)_{{max}} \:=\frac{\mathrm{23}\pi}{\mathrm{6}}\:. \\ $$

Commented by math solver last updated on 09/Feb/18

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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