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Solve for $x$

$\tan x+\sin^2x=\cot^2 x$

I converted everything into $\tan$ but equation is quite complex to solve manually,

$t^5+t^4+t^3=t^2+1$ where $\tan x=t$

Can someone suggest some other method?

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  • $\begingroup$ The rational root theorem fails here and Mathematica does not give a very nice output. I am inclined to think it does not get much simpler than what you have written. $\endgroup$ – Cameron Williams May 18 '16 at 5:52
  • $\begingroup$ Mathematica is suggesting that your polynomial is slightly off, but perhaps it is wrong. It is suggesting that $t$ solves $t^5+t^4-2t^3-t^2+3t-1=0$. $\endgroup$ – Cameron Williams May 18 '16 at 5:55
  • $\begingroup$ I think your polynomial is correct. It only has one real zero, so a) if you can prove that, then b) you can focus efforts on numerical methods for estimating that one value of $t$. $\endgroup$ – alex.jordan May 18 '16 at 6:07
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Given the original equation it is clear that even if there were more than one root you only want positive roots.

Then Descartes' rule of signs applied to $P(t)=t^5+t^4+t^3-t^2-1$ tells you there is exactly one positive root.

Also it is clear that as $P(t)$ changes signs between $0$ and $1$ the root lies between $0$ and $1$. Then applying the bisection method gives the root is at $t\approx 0.87445$.

(The Inverse Symbolic Calculator does not turn up any sensible relations that the root or its inverse tangent, both to 10 digits, satisfy)

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  • $\begingroup$ but according to graph $x=0.719$ is a root and not only this but a lots of solution $\endgroup$ – user5954246 May 18 '16 at 9:35
  • $\begingroup$ Your polynomial gives a solution for $t=tan(x)$ which gives as the first solution for $x=\arctan(0.87445)= 0.71852$. But as $\tan$ is periodic with period $\pi$ there are an infinite number of solutions. $\endgroup$ – Conrad Turner May 18 '16 at 11:31

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