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Express $\theta$ in terms of x if:

$$x+(1/x) = \sqrt{2}\cdot \sec (\theta)$$

A complete solution is always welcome

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closed as off-topic by Jonas Meyer, N. F. Taussig, Mark Fantini, Adam Hughes, 2mkgz Mar 24 at 3:53

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Note that there will be infinitely many $\theta$ for given $x$. It may also be useful to consider $x>0$ and $x<0$ separately. –  André Nicolas Dec 31 '11 at 20:15

2 Answers 2

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  • Square both the sides.

  • Use the fact that $\sec^{2}(\theta) - 1 = \tan^{2}(\theta)$.

Final answer will be:

$$\theta = \tan^{-1}\biggl( \sqrt{\frac{1}{2}\cdot \Bigl(x^{2}+\frac{1}{x^2}\Bigr)}\biggr)$$

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This assumes that $\theta$ is in $(0,\frac{\pi}{2})$. Why not $\theta=\cos^{-1}\left(\frac{x\sqrt 2}{x^2+1}\right)$, which will at least give the correct answer for $\theta$ in $(0,\pi)$? (It also doesn't involve squaring or taking square roots or using a Pythagorean identity.) –  Jonas Meyer Dec 31 '11 at 20:25

The function $f(x)=x+{1\over x}$ has a minimum value of $2$ for $x>0$ and a maximum value of $-2$ for $x<0$. Thus $|f(x)/\sqrt 2| \ge \sqrt 2\ge 1$ for all $x\ne0$, and $$ \theta=\sec^{-1}\Bigr(\, {\textstyle{1\over\sqrt2}} (x+{\textstyle{1\over x}})\,\Bigr) $$ where $\sec^{-1}$ is the inverse function for $\sec$ restricted to $[0,\pi]\setminus\{\pi/2\}$ (so $\sec^{-1}$ has domain $|x|\ge 1$).

Of course, as Andre points out in the comments, this isn't a unique solution.

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