Trigonometric inequation with tangent function

[karseme]

I just want your opinion on my attempt at a solution of this task:

\(\displaystyle \tan{\dfrac{x}{2}}>\dfrac{\tan{x}-2}{\tan{x}-2}\)

My attempt:

We know that:

\(\displaystyle \tan{x}=\dfrac{2\tan{\dfrac{x}{2}}}{1-\tan^2{\dfrac{x}{2}}} \)

But, at the beginning we should set limits to tangent function:

\(\displaystyle \dfrac{x}{2} \neq \dfrac{\pi}{2}+k\pi, k\in\mathbb{Z} \qquad x \neq \dfrac{\pi}{2}+k\pi, k\in\mathbb{Z} \\
x \neq \pi + 2k\pi,k\in\mathbb{Z} \qquad x \neq \dfrac{\pi}{2}+k\pi, k\in\mathbb{Z} \)

If we use the identity that I've given above, we get:

\(\displaystyle \tan{\dfrac{x}{2}} > \dfrac{2\tan^2{\dfrac{x}{2}}+2\tan{\dfrac{x}{2}}-2}{2+2\tan{\dfrac{x}{2}}-2\tan^2{\dfrac{x}{2}}} \)

Now we can have two cases: denominator is less than zero and denominator is larger than zero.
Let's use substitution \(\displaystyle \tan{\dfrac{x}{2}}=m \) for the convenience, then we have quadratic equation:

\(\displaystyle -2m^2+2m+2=0 \), with the solutions \(\displaystyle m_{1,2}=\dfrac{1 \pm \sqrt{5}}{2} \).
Now, denominator is larger than 0 for \(\displaystyle m \in \langle \dfrac{1 - \sqrt{5}}{2},\dfrac{1 + \sqrt{5}}{2} \rangle \implies \tan{\dfrac{x}{2}}\in\langle \dfrac{1 - \sqrt{5}}{2},\dfrac{1 + \sqrt{5}}{2} \rangle \).
Now we can multiply inequality with denominator because denominator is not 0 for those values and it is positive so inequality sign won't "change", we have:

\(\displaystyle \tan{\dfrac{x}{2}}(2+2\tan{\dfrac{x}{2}}-2\tan^2{\dfrac{x}{2}}) > 2\tan^2{\dfrac{x}{2}}+2\tan{\dfrac{x}{2}}-2 \)

We have:
\(\displaystyle \tan^3{\dfrac{x}{2}} < 1 \implies \tan{\dfrac{x}{2}} < 1 \)

Then,

\(\displaystyle \tan{\dfrac{x}{2}} \in \langle \dfrac{1 - \sqrt{5}}{2},1 \rangle \\ \dfrac{x}{2} \in \langle \arctan{\dfrac{1 - \sqrt{5}}{2}}+k\pi, \dfrac{\pi}{4}+k\pi \rangle ,k\in\mathbb{Z} \\ x \in \langle 2\arctan{\dfrac{1 - \sqrt{5}}{2}}+2k\pi, \dfrac{\pi}{2}+2k\pi \rangle ,k\in\mathbb{Z} \)
Which is a good solution because it does not contain our limits for tangent function.

Now for the second case, we assume that: \(\displaystyle m \in \langle -\infty,\dfrac{1 - \sqrt{5}}{2} \rangle \cup \langle \dfrac{1 + \sqrt{5}}{2}, +\infty \rangle \implies \tan{\dfrac{x}{2}}\in \langle -\infty,\dfrac{1 - \sqrt{5}}{2} \rangle \cup \langle \dfrac{1 + \sqrt{5}}{2}, +\infty \rangle\)

Now we can multiply the inequality with the denominator but the ">" sign will "change":

\(\displaystyle \tan{\dfrac{x}{2}}(2+2\tan{\dfrac{x}{2}}-2\tan^2{\dfrac{x}{2}}) < 2\tan^2{\dfrac{x}{2}}+2\tan{\dfrac{x}{2}}-2 \)

We have:

\(\displaystyle \tan^3{\dfrac{x}{2}} > 1 \implies \tan{\dfrac{x}{2}} > 1 \)

Then,

\(\displaystyle \tan{\dfrac{x}{2}} \in \langle \dfrac{1 + \sqrt{5}}{2}, +\infty \rangle \\ \dfrac{x}{2} \in \langle \arctan{\dfrac{1 + \sqrt{5}}{2}}+k\pi, \dfrac{\pi}{2}+k\pi \rangle ,k\in\mathbb{Z} \\ x \in \langle 2\arctan{\dfrac{1 + \sqrt{5}}{2}}+2k\pi, \pi+2k\pi \rangle ,k\in\mathbb{Z} \)
Which is a good solution because it does not contain our limits for tangent function.

And finally the solution is:

\(\displaystyle x \in \langle 2\arctan{\dfrac{1 - \sqrt{5}}{2}}+2k\pi, \dfrac{\pi}{2}+2k\pi \rangle \cup \langle 2\arctan{\dfrac{1 + \sqrt{5}}{2}}+2k\pi, \pi+2k\pi \rangle ,k\in\mathbb{Z} \)

Is this a good solution or? If not please help me solve this task. Appreciate your help.

 

[jvicens]

I appreciate the effort you put into finding a solution for this task. However, I have some concerns about your approach.

Firstly, it is not necessary to set limits for the tangent function at the beginning. The inequality \tan{\dfrac{x}{2}} > \dfrac{\tan{x}-2}{\tan{x}-2} is already defined for all values of x except when \tan{x} = 2. So, your limits are not relevant to the problem at hand.

Secondly, your solution only considers the case when the denominator is positive. You should also consider the case when the denominator is negative. In this case, the inequality sign should be reversed, and your solution will be different.

Lastly, your final solution is not complete. You have not considered the case when \tan{x} = 2. In this case, the inequality is not defined, and your solution does not cover this scenario.

In conclusion, your attempt at a solution has some flaws and is not complete. I suggest double-checking your approach and considering all possible cases before presenting a solution. Keep up the good work and continue to improve your problem-solving skills.