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I am trying to rearrange the following formula to make x the subject. $\frac{x}{180^\circ}\cdot \pi - \sin(x)\cos(x) = \frac{\pi}{y}$. I appreciate that I could use the double angle identity in reverse, but this still isn't getting me anywhere. This is not for school, I am playing with an idea and the math is getting a bit above me. Thank you

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  • $\begingroup$ Welcome to MSE. Please use MathJax to format your questions. $\endgroup$ – user507623 Dec 26 '17 at 9:55
  • $\begingroup$ you can convert $\sin(x),\cos(x)$ into $\tan(\frac{x}{2})$ $\endgroup$ – Dr. Sonnhard Graubner Dec 26 '17 at 9:59
  • $\begingroup$ The simplest way is, perhaps, to put $\;\sin x\cos x=\frac12\sin2x\;$ ...Still, you get an equation in which you won't be able to "separate" $\;x\;$ by means of elementary functions. $\endgroup$ – DonAntonio Dec 26 '17 at 10:01
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When doing maths at school, we go through a lot of exercises where we are asked to rearrange the expression so that one variable is expressed as a formula of the other(s). Probably this gives an impression that it is always possible (and if you cannot do it, it is because the "math is above you" etc.)

Do not get discouraged. It is not always possible. Actually, it is very rarely possible. Those school examples where it is possible are just carefully chosen precisely so that it is possible; for a random formula with $x $ and $y $ it would be a surprise to be able to extract one of the variables as a function of the other.

A historical note: look up the problem of solving equations in radicals. You may know that, if you have $y=ax^2+bx+c $, you can solve it for $x $ as a quadratic equation and get the solution expressed using the square root... People have solved the same for a cubic and quartic expression in $x $, and then got stuck for centuries trying to solve the 5th degree equation (in effect solve $y=ax^5+bx^4+cx^3+dx^2+ex+f $ for $x $). In the 19th century, based on the work of Galois, Abel, Gauss and others, it was proven that it is not possible (at least not using roots), but not before this work created foundations for completely new (at the time) branches of algebra such as group theory and field theory.

So that is how hard the things can go. For us mortals: try a bit longer, but don't spend too much time. Or be prepared to invent new maths, if you are inclined so.

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  • $\begingroup$ Thankyou! How fascinating, although quite frustrating that it now means I can't just find a useful formula for what I'm trying to solve. If you don't mind me asking, do you know how the graphing calculators find a solution to this, then? Am I correct in therefore assuming that they get an approximate answer by using some advanced sort of trial and improvement technique (such as the Newton-Raphson method, or an iterative formula?) $\endgroup$ – Purple dragon unicorn Dec 26 '17 at 11:01
  • $\begingroup$ You are spot on: if you cannot solve the equation analytically, you can still solve it numerically, using one of the algorithms you've mentioned, or many others. $\endgroup$ – user491874 Dec 26 '17 at 11:04

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