tibukooinm

2022-03-25

For which values a, the equation
$a\mathrm{sin}x+\left(a+1\right){\mathrm{sin}}^{2}\frac{x}{2}+\left(a-1\right){\mathrm{cos}}^{2}\frac{x}{2}=1$

Leonardo Mcpherson

$a\mathrm{sin}x+\left(a+1\right){\mathrm{sin}}^{2}\frac{x}{2}+\left(a-1\right){\mathrm{cos}}^{2}\frac{x}{2}=1$
$a\mathrm{sin}x+a\left({\mathrm{sin}}^{2}\frac{x}{2}+{\mathrm{cos}}^{2}\frac{x}{2}\right)-\left({\mathrm{cos}}^{2}\frac{x}{2}-{\mathrm{sin}}^{2}\frac{x}{2}\right)=1$
$a\mathrm{sin}x+a-\mathrm{cos}x=1$
$a\mathrm{sin}x-\mathrm{cos}x=\left(1-a\right)$
$\frac{a}{\sqrt{1+{a}^{2}}}\mathrm{sin}x-\frac{1}{\sqrt{1+{a}^{2}}}\mathrm{cos}x=\frac{\left(1-a\right)}{\sqrt{1+{a}^{2}}}$
$\mathrm{sin}x\mathrm{cos}\alpha -\mathrm{cos}\mathrm{sin}\alpha =\mathrm{sin}\beta$
Where $\alpha ={\mathrm{cos}}^{-1}\left(\frac{a}{\sqrt{1+{a}^{2}}}\right)$ and $\beta ={\mathrm{sin}}^{-1}\left(\frac{1-a}{\sqrt{1+{a}^{2}}}\right)$
$\mathrm{sin}\left(x-\alpha \right)=\mathrm{sin}\beta$
$x-\alpha =2n\pi +\beta ⇒x=2n\pi +\alpha +\beta$
or
$x-\alpha =\left(2n+1\right)\pi -\beta ⇒x=\left(2n+1\right)\pi +\alpha -\beta$

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