I come up with the equation:

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?

  • 4
    $\begingroup$ There is no simple solution. You should use numerical methods. $\endgroup$ – Math-fun Mar 31 '16 at 10:49
  • $\begingroup$ 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). $\endgroup$ – Claude Leibovici Mar 31 '16 at 10:51
  • $\begingroup$ 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. $\endgroup$ – Mc Cheng Mar 31 '16 at 10:57
  • $\begingroup$ WA doesn't seem to be finding closed forms for the solutions. $\endgroup$ – zz20s Mar 31 '16 at 10:59
  • $\begingroup$ Plot it on Wolfram Alpha graphing... $\endgroup$ – tatan Apr 3 '16 at 15:25

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


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

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.

  • $\begingroup$ Note the denominator in the RHS is $\sec^2(*2x*)$. $\endgroup$ – Oscar Lanzi Apr 3 '16 at 16:14
  • $\begingroup$ omg! , thanks shall correct. $\endgroup$ – Narasimham Apr 3 '16 at 16:56

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