# Trigonometric equation $\cot(x)\csc(x)=2\sec^2(2x)$

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?

• 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... – tatan Apr 3 '16 at 15:25

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.

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