What am I doing wrong with solving $2\tanh^2x-\text{sech}~x=1$?

Question

$2\tanh^2(x)-\text{sech}(x)=1$

$\tanh^2(x)=1-\text{sech}^2(x)$

$2(1-\text{sech}^2(x))-\text{sech}(x)=1$

$2\text{sech}^2(x)+\text{sech}(x)-1=0$

$\text{sech}(x)=\frac{1}{2} $ Not possible. And $\text{sech}(x)=-1$ Also not possible

What am I doing wrong?

score 3 · Accepted Answer · answered Jun 01 '13 at 00:29

3

The equation $\text{sech} x=\frac{1}{2}$ has a solution.

Rewrite as $\cosh x=2$, and then as $e^x+e^{-x}=4$, and then as $e^{2x}-4e^{x}+1=0$. We get a reasonably pleasant quadratic in $e^x$.

answered Jun 01 '13 at 00:29

André Nicolas

507,029

$\frac{1}{cosh^{-1}(\frac{1}{2})}$ Doesn't? – maxmitch Jun 01 '13 at 00:31
1

I have added how to solve it. – André Nicolas Jun 01 '13 at 00:32
I suck at algebra... ha. Thanks! – maxmitch Jun 01 '13 at 00:33
Is there only one solution? $ln(2+\sqrt{3})$ The mark scheme says $-ln(2+\sqrt{3})$ as well! Why is that? – maxmitch Jun 01 '13 at 00:35
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What about $\ln(2-\sqrt{3})$? Since $\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$, this agrees with the marking scheme. Anyway, $\cosh $ is symmetric about the $y$-axis, so if $x_0$ is a solution so is $-x_0$. – André Nicolas Jun 01 '13 at 00:37
I have been given $cosh^{-1}(x)=ln(x+\sqrt{x^2-1})$ – maxmitch Jun 01 '13 at 00:39
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Well, that just short-circuits the $e^x$ calculation that I made. – André Nicolas Jun 01 '13 at 00:40
I don't understand your last comment. I think I need to learn what hyperbolic graphs look like! The question hasn't given a range so surely there are an infinite number of solutions? – maxmitch Jun 01 '13 at 00:43
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There are precisely two. The function $\cosh$ decreases until $x=0$, then increases. Doesn't wiggle like $\cos$. – André Nicolas Jun 01 '13 at 00:45
Ohhh sweet thanks! I wish you could draw hyperbolic graphs on a graphical calculator! – maxmitch Jun 01 '13 at 00:46
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As to the comment, let $y=\cosh x=\frac{e^x +e^{-x}}{2}$. We want to solve $\cosh x=y$. Using same technique as in my answer, we get $e^{2x}-2ye^{x}+1=0$, so $e^x=y\pm\sqrt{y^2-1}$. Interchange the roles of $x$ and $y$, and remember that $\cosh^{-1} x$ is the positive number whose cosh is $x$. – André Nicolas Jun 01 '13 at 00:49
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About graphing calculator, I am sure you can. If it doesn't know $\cosh x$, tell it to graph $\frac{e^x+e^{-x}}{2}$. – André Nicolas Jun 01 '13 at 00:54

What am I doing wrong with solving $2\tanh^2x-\text{sech}~x=1$?

1 Answers1