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I come up with the equation:
$\cot(x)\csc(x)=2\sec^2(2x)$

This looks extremely simple but I am not able to come up with a simple solution for $x$.
I work out that $4\cos^5(x)-4\cos^3(x)+2\cos^2(x)+\cos(x)-2=0$
Am I on the right track or is there an easier approach to solving the equation?

Mc Cheng
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    There is no simple solution. You should use numerical methods. – Math-fun Mar 31 '16 at 10:49
  • Your equation is fine. Define $y=\cos(x)$ and plot your function for $-1\leq y\leq 1$. You will see only one root. And, as Math-fun commented, use some numerical method (Newton would work very ewell). – Claude Leibovici Mar 31 '16 at 10:51
  • Of course I know the graphing method and Newton method for solving the quintic equations. I doubt if some trigonometric identities can be applied to the above equation. – Mc Cheng Mar 31 '16 at 10:57
  • WA doesn't seem to be finding closed forms for the solutions. – zz20s Mar 31 '16 at 10:59
  • Plot it on Wolfram Alpha graphing... – Soham Apr 03 '16 at 15:25
  • And I tested explicitly for quadratic factors (the leading coefficient, constant term, and sum of all coefficients must divide the corresponding quantities in the quintic polynomial; only a finite number of candidates survive these constraints). Found there are none. And of course candidate rational roots do not work either. So we have an irreducible quintic, bring on the numerical/graphical solution. – Oscar Lanzi Apr 03 '16 at 19:06

1 Answers1

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Since my school days I used to use a technique to start it all:

EDIT 1:

$$ \frac{c}{s} \cdot \frac1s = \frac{2}{(2 c^2 -1)^2 }$$

Simplifying using $ s^2 = (1-c^2) $ we are both getting

$$ 4 c^5 + 0 c^4 - 4 c^3 + 2 c^2 + c -2 = 0\, $$

$ c = 1,\, -1 $ are not roots. Only numerical iteration methods e.g. Newton would help in evaluating the single real root.$ \cos(x)\approx 0.9042086 $ is a real root, four others are imaginary as per WA or any CAS plot of the above $5^{th}$ degree polynomial.

Narasimham
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