# A calculus variable conversion question [closed]

Express $\theta$ in terms of x if:

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

A complete solution is always welcome

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, N. F. Taussig, Mark Fantini, Adam Hughes, 2mkgz
If this question can be reworded to fit the rules in the help center, please edit the question.

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

Hints:

• Square both the sides.

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

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