While practicing trigonometry online I came across this one exercise which was too hard for me to solve:

$$ (\tan{4x} + \cot{4x}-\frac{5}{2})(\cos(2x) - |\sin{x}|) > 0 $$

In my humble opinion this is kind of overkill, because it exceeds even WolframAlpha's computation time and even then the solutions are uncommon values.

Frankly I don't even know where to start, but I'd start by noticing this. Let's say we have two expressions divided by brackets. Our condition will be true if either both of them are $>0$ or both of them are $<0$.

Now where I would have trouble is taking the intersection of all those sets of real numbers, so I was thinking that there must be an easier way to do this.

Can anyone help?

  • $\begingroup$ You can say $x = \frac{\arctan{y}}{4}$, then you would get equations involving $y$, can you try this $\endgroup$ Mar 20 '21 at 7:14


  • For the first term set $t=\tan(4x)$ and you get a quadratic in $t$ to solve.
  • Then solve $\tan(4x)=t$ for the $2$ roots found.
  • For the second one set $s=\sin(x)$ and rewrite $\cos(2x)$ in function of $s^2$ another quadratic (well two in fact because of absolute value) to solve.
  • Then solve $\sin(x)=s$ for the $4$ roots found.
  • Find the minimal interval for describing the solution and extend all solutions by periodicity


$(\tan{4x} + \cot{4x}-\frac{5}{2})(\cos(2x) - |\sin{x}|) > 0$

$(\tan{4x} + \frac{1}{\tan{4x}}-\frac{5}{2})$$(1-2|sinx|^2 - |\sin{x}|)$ > 0

A= $\tan{4x} + \frac{1}{\tan{4x}}-\frac{5}{2}$

B=$1-2|sinx|^2 - |\sin{x}|$

So either (A <0 and B<0) or (A>0, B>0) , we can continue from here...


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