5
$\begingroup$

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?

$\endgroup$
  • 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.