DF equation: solve for x

Question

OK, i know what to do with the first and second term. Could someone please help with the third and solve for x? :) [This might seem a silly question though.]

Answer 1

Put $x=tan(\frac{1}{2}\alpha)$. Then:

$\large \frac{2x}{1+x^2}= \frac{2tan(\frac{1}{2}\alpha)}{1+tan^2(\frac{1}{2}\alpha)}= \frac{2cos(\frac{1}{2}\alpha)sin(\frac{1}{2}\alpha)}{cos^2(\frac{1}{2}\alpha)+sin^2(\frac{1}{2}\alpha)}=\frac{sin(\alpha)}{1}= \normalsize sin(\alpha)$.

And

$\large \frac{1-x^2}{1+x^2}= \frac{1-tan^2(\frac{1}{2}\alpha)}{1+tan^2(\frac{1}{2}\alpha)}= \frac{cos^2(\frac{1}{2}\alpha)-sin^2(\frac{1}{2}\alpha)}{cos^2(\frac{1}{2}\alpha)+sin^2(\frac{1}{2}\alpha)}=\frac{cos(\alpha)}{1} = \normalsize cos(\alpha)$.

Plugging this into the equation we get:

$3sin^{-1}(sin(\alpha))-cos^{-1}(cos(\alpha))+2tan^{-1}(sin(\alpha))=\pi/3 \Leftrightarrow$

$3\alpha-\alpha +2tan^{-1}(sin(\alpha))=\pi/3 \Leftrightarrow$

$sin(\alpha)=tan(\pi/6-\alpha)$. $(*)$

From this point you have to solve this equation numerically (for example use this graphical calculator).

For example $\alpha=0.266284..$ is a solution and hence $x=0.13393434..$ is an approximate solution. Of course I am a bit imprecise, since there are restrictions to the solution of $(*)$ (for example $-1 \leq tan(\pi/6-\alpha) \leq 1$) and taking periodicity of trigoniometric functions into account. I leave that to you.