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solve-sin-2-x-sin-2-y-1-sinx-siny-sinxsiny-




Question Number 165876 by bounhome last updated on 09/Feb/22
solve:    sin^2 x+sin^2 y+1=sinx+siny+sinxsiny
$${solve}: \\ $$$$\:\:{sin}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {y}+\mathrm{1}={sinx}+{siny}+{sinxsiny} \\ $$
Commented by MJS_new last updated on 10/Feb/22
u=sin x ∧v=sin y  u^2 +v^2 +1=u+v+uv  ⇔  v^2 −(u+1)v+u^2 −u+1=0  v∈R ⇔ −(3/4)(u−1)^2 ≥0 which is only possible with u=1  ⇒ v=−1  ⇒ sin x =1∧sin y =−1
$${u}=\mathrm{sin}\:{x}\:\wedge{v}=\mathrm{sin}\:{y} \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} +\mathrm{1}={u}+{v}+{uv} \\ $$$$\Leftrightarrow \\ $$$${v}^{\mathrm{2}} −\left({u}+\mathrm{1}\right){v}+{u}^{\mathrm{2}} −{u}+\mathrm{1}=\mathrm{0} \\ $$$${v}\in\mathbb{R}\:\Leftrightarrow\:−\frac{\mathrm{3}}{\mathrm{4}}\left({u}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{with}\:{u}=\mathrm{1} \\ $$$$\Rightarrow\:{v}=−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\mathrm{1}\wedge\mathrm{sin}\:{y}\:=−\mathrm{1} \\ $$
Commented by mindispower last updated on 10/Feb/22
u=1
$${u}=\mathrm{1} \\ $$
Commented by MJS_new last updated on 10/Feb/22
yes.
$$\mathrm{yes}. \\ $$

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