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How do I express k in terms of x in the following equation:

$$x = \frac{sin(4k)}{4} + k$$

Thank you.

[Edit]

If you're wondering, this is not from a homework question. I was playing around with sin and cos functions, and got to the gradient of each of these. I tried to find the general equation for the intersection between the function of the two gradients, but could only find the parametric form of the equation. The equation for x here is the parametric form for the x-coordinate of the points of intersection of the gradient functions. I genuinely don't know how to solve this. (

[Edit 2] In case you want the full detail. I've started with two simple $y = tan\ x$ and $y = cot\ x$ functions (sorry for the mistake in the first edit). The gradients of these are $\frac{dy}{dx} = sec^2\ x$ and $\frac{dy}{dx} = -cosec^2 \ x$ respectively. Therefore the general formulas for the equation of tangents are: $$ y = (sec\ x)^2(x-k) + tan\ k$$ and $$y = (cosec \ x)^2(x-k) + cot\ k$$ where k is any real number, respectively.

By equating the two equations, I was able to solve for the x-value at which the two gradients intersect. $$x = (cot\ k - tan\ k)(sin(k)cos(k))^2 + k$$ which then simplifies to $$x = \frac{sin\ 4k}{4} + k$$

Also, by substituting x into the original gradient equations, I was able to find the y-value for the points of intersection between the two gradients. $$y = (cot\ k - tan\ k)(sin\ k)^2 + tan\ k= (tan\ k - cot\ k)(cos\ k)^2 + cot\ k$$

This simplifies to: $$y= \frac{1}{2}(1 - tan^2\ k)(sin\ 2k) + tan\ k = sin\ 2k$$

(I might have made multiple mistakes so far...)

The point is I was trying to find k in terms of x so that I could find y in terms of x, instead of leaving things in parametric form.

Here is the graph of the parametric function I've plotted, representing the points of intersection of the two gradients I've mentioned:

enter image description here

[Edit 3]

I got it, it wasn't so obvious when I didn't see I could simplify some equations. Thanks.

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  • $\begingroup$ Evidently $k\approx x$ $\endgroup$ – Minz Jun 9 '17 at 1:35
  • $\begingroup$ True, but I need the exact solution, since I'm using the solution to k for expressing my parametric equation as cartesian. $\endgroup$ – user147526 Jun 9 '17 at 1:41
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    $\begingroup$ the exact solution is impossible in analitic form $\endgroup$ – Minz Jun 9 '17 at 1:47
  • $\begingroup$ If you edit the question to give the equations from which you got this one (it's not clear to me which gradients you mean), perhaps someone will have a useful insight. $\endgroup$ – David K Jun 9 '17 at 1:53
  • $\begingroup$ Thanks, I've already edited it. $\endgroup$ – user147526 Jun 9 '17 at 2:37

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