parksinta8rkv

2022-03-23

Solve this equation.
${\left(\sqrt{\sqrt{2}+1}\right)}^{\mathrm{sin}\left(x\right)}+{\left(\sqrt{\sqrt{2}-1}\right)}^{\mathrm{sin}\left(x\right)}=2$

Ashley Olson

To solve
${\left(\sqrt{\sqrt{2}+1}\right)}^{\mathrm{sin}\left(x\right)}+{\left(\sqrt{\sqrt{2}-1}\right)}^{\mathrm{sin}\left(x\right)}=2$
let us rewrite as
${\left(\sqrt{2}+1\right)}^{\frac{1}{2}\mathrm{sin}\left(x\right)}+{\left(\sqrt{2}-1\right)}^{\frac{1}{2}\mathrm{sin}\left(x\right)}=2$
First consider the auxiliary equation
$y+\frac{1}{y}=2$
You can verify this has only one solution, $y=1$. Next, notice
$\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$
By laws of exponents, our original equation is then
${\left(\sqrt{2}+1\right)}^{\frac{1}{2}\mathrm{sin}\left(x\right)}+\frac{1}{{\left(\sqrt{2}+1\right)}^{\frac{1}{2}\mathrm{sin}\left(x\right)}}=2$
which we recognize as the auxiliary equation with $y={\left(\sqrt{2}+1\right)}^{\frac{1}{2}\mathrm{sin}\left(x\right)}$. Therefore the solution to our original equation has
${\left(\sqrt{2}+1\right)}^{\frac{1}{2}\mathrm{sin}\left(x\right)}=1$
Since $\sqrt{2}+1>1$, we must have
$\frac{1}{2}\mathrm{sin}\left(x\right)=0$
The solutions are the integer multiples of $\pi$

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