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Question Number 15298 by Tinkutara last updated on 09/Jun/17
Solve for x and y x^2 + 2x sin(xy) + 1 = 0
$$\mathrm{Solve}\:\mathrm{for}\:{x}\:\mathrm{and}\:{y} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:\mathrm{sin}\left({xy}\right)\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$
Commented by prakash jain last updated on 10/Jun/17
xy=u x^2 +2xsin (u)+1=0 sin u=−((1+x^2 )/(2x)) f(x)=−((1+x^2 )/(2x)) f(x)≤−1 or f(x)≥1 only 2 solution for x for f(x)=±1 f(x)=−1⇒x=1 f(x)=1⇒x=−1 x=1 sin xy=−((1+x^2 )/(2x))=−1 sin (y)=−1⇒y=−(π/2) x=−1 sin (−y)=−((1+x^2 )/(2x)) sin (−y)=1 sin (y)=−1⇒sin y=−1⇒y=−(π/2) solutions: x=1,y=−(π/2) (y can take 2nπ+((3π)/2)) x=−1,y=−(π/2) (y can take 2nπ+((3π)/2))
$${xy}={u} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}\mathrm{sin}\:\left({u}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:{u}=−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$${f}\left({x}\right)\leqslant−\mathrm{1}\:{or}\:{f}\left({x}\right)\geqslant\mathrm{1} \\ $$$${only}\:\mathrm{2}\:{solution}\:{for}\:{x}\:{for}\:{f}\left({x}\right)=\pm\mathrm{1} \\ $$$${f}\left({x}\right)=−\mathrm{1}\Rightarrow{x}=\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{1}\Rightarrow{x}=−\mathrm{1} \\ $$$${x}=\mathrm{1} \\ $$$$\mathrm{sin}\:{xy}=−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}}=−\mathrm{1} \\ $$$$\mathrm{sin}\:\left({y}\right)=−\mathrm{1}\Rightarrow{y}=−\frac{\pi}{\mathrm{2}} \\ $$$${x}=−\mathrm{1} \\ $$$$\mathrm{sin}\:\left(−{y}\right)=−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$$\mathrm{sin}\:\left(−{y}\right)=\mathrm{1} \\ $$$$\mathrm{sin}\:\left({y}\right)=−\mathrm{1}\Rightarrow\mathrm{sin}\:{y}=−\mathrm{1}\Rightarrow{y}=−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{solutions}: \\ $$$${x}=\mathrm{1},{y}=−\frac{\pi}{\mathrm{2}}\:\left({y}\:{can}\:{take}\:\mathrm{2}{n}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$$${x}=−\mathrm{1},{y}=−\frac{\pi}{\mathrm{2}}\:\left({y}\:{can}\:{take}\:\mathrm{2}{n}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$
Commented by Tinkutara last updated on 10/Jun/17
But answer is (x, y) = (1, (2n + 1)((3π)/2)) and (−1, (2n + 1)(π/2))
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\left({x},\:{y}\right)\:= \\ $$$$\left(\mathrm{1},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:\mathrm{and}\:\left(−\mathrm{1},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right) \\ $$
Commented by prakash jain last updated on 10/Jun/17
x^2 +2xsin (xy)+1=0 put x=1 2sin y+2=0 sin y=−1 what is the solution for y. compare it with ur book′s answer.
$${x}^{\mathrm{2}} +\mathrm{2}{x}\mathrm{sin}\:\left({xy}\right)+\mathrm{1}=\mathrm{0} \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\mathrm{2sin}\:{y}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{sin}\:{y}=−\mathrm{1} \\ $$$${what}\:{is}\:{the}\:{solution}\:{for}\:{y}. \\ $$$${compare}\:{it}\:{with}\:{ur}\:{book}'{s}\:{answer}. \\ $$
Commented by prakash jain last updated on 10/Jun/17
3nπ+((3π)/2) n=0 y=((3π)/2) n=1 y=3π+((3π)/2)=3π+π+(π/2)=4π+(π/2) put x=1,y=4π+(π/2) in original equation does it satisfy the original equation?
$$\mathrm{3}{n}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}}\: \\ $$$${n}=\mathrm{0}\:{y}=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${n}=\mathrm{1}\:{y}=\mathrm{3}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}}=\mathrm{3}\pi+\pi+\frac{\pi}{\mathrm{2}}=\mathrm{4}\pi+\frac{\pi}{\mathrm{2}} \\ $$$${put}\:{x}=\mathrm{1},{y}=\mathrm{4}\pi+\frac{\pi}{\mathrm{2}}\:\mathrm{in}\:\mathrm{original}\:\mathrm{equation} \\ $$$$\mathrm{does}\:\mathrm{it}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{original}\:\mathrm{equation}? \\ $$
Answered by Tinkutara last updated on 09/Jul/17
Given equation can be written as [x + sin (xy)]^2 + cos^2 (xy) = 0 ∴ cos (xy) = 0 and x + sin (xy) = 0. ⇒ x = ± 1 cos y = 0; sin y = −1 ⇒ y = (2n + 1)((3π)/2) ⇒ (x, y) = (± 1, (2n + 1)((3π)/2))
$$\mathrm{Given}\:\mathrm{equation}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$$\left[{x}\:+\:\mathrm{sin}\:\left({xy}\right)\right]^{\mathrm{2}} \:+\:\mathrm{cos}^{\mathrm{2}} \:\left({xy}\right)\:=\:\mathrm{0} \\ $$$$\therefore\:\mathrm{cos}\:\left({xy}\right)\:=\:\mathrm{0}\:\mathrm{and}\:{x}\:+\:\mathrm{sin}\:\left({xy}\right)\:=\:\mathrm{0}. \\ $$$$\Rightarrow\:{x}\:=\:\pm\:\mathrm{1} \\ $$$$\mathrm{cos}\:{y}\:=\:\mathrm{0};\:\mathrm{sin}\:{y}\:=\:−\mathrm{1}\:\Rightarrow\:{y}\:=\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\left({x},\:{y}\right)\:=\:\left(\pm\:\mathrm{1},\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$

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