Question Number 29478 by math solver last updated on 09/Feb/18
$${the}\:{number}\:{of}\:{ordered}\:{pairs}\:\left({x},{y}\right) \\ $$$${of}\:{real}\:{numbers}\:{satisfying}\: \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}={sin}^{\mathrm{2}} {y} \\ $$$${and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\:\mathrm{3}\:{is}\:? \\ $$
Answered by mrW2 last updated on 09/Feb/18
$$\mathrm{4}\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{1}={sin}^{\mathrm{2}} {y} \\ $$$$\mathrm{4}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}={sin}^{\mathrm{2}} {y} \\ $$$${LHS}\geqslant\mathrm{1} \\ $$$${RHS}\leqslant\mathrm{1} \\ $$$$\Rightarrow{LHS}={RHS}=\mathrm{1} \\ $$$$\Rightarrow{x}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:{y}=\pm\mathrm{1}\Rightarrow{y}=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}+{y}^{\mathrm{2}} \leqslant\:\mathrm{3} \\ $$$$\Rightarrow{y}^{\mathrm{2}} <\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$\Rightarrow−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}<{y}<\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\pm\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},−\frac{\pi}{\mathrm{2}}\right)\:{and}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 09/Feb/18
w♡w мяω2!
Commented by math solver last updated on 09/Feb/18
$${thank}\:{you}\:{sir}. \\ $$
Commented by NECx last updated on 09/Feb/18
$${wow}...\:{Thanks}\:{sir} \\ $$