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Question Number 184622 by Shrinava last updated on 09/Jan/23

Solve for real numbers: sinx (√(1 − sin^2 x)) = 1 + cosy (√(1 − cos^2 y))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{sinx}\:\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:=\:\mathrm{1}\:+\:\mathrm{cosy}\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}^{\mathrm{2}} \mathrm{y}}\: \\ $$

Answered by floor(10²Eta[1]) last updated on 09/Jan/23

sinxcosx=1+senycosy ((sen(2x))/2)=1+((sen(2y))/2) sen(2x)=2+sen(2y) −1≤sen(2x)≤1 1≤2+sen(2y)≤3 ⇒sen2x=1=2+sen2y ⇒sen2x=1 ∧ sen2y=−1 2x=(π/2)+2nπ ∧ 2y=((3π)/2)+2mπ ⇒x=(π/4)+nπ ∧ y=((3π)/4)+mπ, m,n∈Z

$$\mathrm{sinxcosx}=\mathrm{1}+\mathrm{senycosy} \\ $$$$\frac{\mathrm{sen}\left(\mathrm{2x}\right)}{\mathrm{2}}=\mathrm{1}+\frac{\mathrm{sen}\left(\mathrm{2y}\right)}{\mathrm{2}} \\ $$$$\mathrm{sen}\left(\mathrm{2x}\right)=\mathrm{2}+\mathrm{sen}\left(\mathrm{2y}\right) \\ $$$$−\mathrm{1}\leqslant\mathrm{sen}\left(\mathrm{2x}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{1}\leqslant\mathrm{2}+\mathrm{sen}\left(\mathrm{2y}\right)\leqslant\mathrm{3} \\ $$$$\Rightarrow\mathrm{sen2x}=\mathrm{1}=\mathrm{2}+\mathrm{sen2y} \\ $$$$\Rightarrow\mathrm{sen2x}=\mathrm{1}\:\wedge\:\mathrm{sen2y}=−\mathrm{1} \\ $$$$\mathrm{2x}=\frac{\pi}{\mathrm{2}}+\mathrm{2n}\pi\:\wedge\:\mathrm{2y}=\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{2m}\pi \\ $$$$\Rightarrow\mathrm{x}=\frac{\pi}{\mathrm{4}}+\mathrm{n}\pi\:\wedge\:\mathrm{y}=\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{m}\pi,\:\mathrm{m},\mathrm{n}\in\mathbb{Z} \\ $$

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