I am trying to find a way to find the self intersection point of a certain problem. The problem is:
$$\begin{cases} x = 1-4\cos^2(t) \\ y = \tan(t)(1-4\cos^2(t)) \\ -\tfrac\pi2 < t < \tfrac\pi2 \end{cases}$$
I was trying to use the methods taught in my calculus II class, however, I am having a bit of difficulty. I tried doing something like: $x(t_1) = x(t_2)$, and then doing the same for y. The answer that I am getting is $t = 0$, however, I know that $t_1 \neq t_2$. I am asking for any assistance/ guidance in finding the correct solution. Any help is appreciated! This is what I have done:
$$\begin{align*} 1-4\cos^2(t_1) &= 1- 4\cos^2(t_2) \\ \cos^2(t_1) &= \cos^2(t_2) \\ t_1 &= -t_2 \end{align*}$$
Since $\cos(t)$ will be the same in Q(IV) and Q(I), $\cos(-t) = \cos(t)$,
$$\begin{align*} \tan(t_1)(1-4\cos^2(t_1)) &= \tan(-t_1)(1-4\cos^2(t_1)) \\ \tan(t_1) &= \tan(-t_1) \\ t_1 &= 0 \text{ or } \pi \end{align*}$$
however $\pi$ is outside the domain.
From the equations $x=1-4\cos(t)^2, y=x\tan(t)$, for a self-intersection at $t_1$ and $t_2$ you must have $\cos(t_1)^2=\cos(t_2)^2$. And, if $x\ne0$ at the self-intersection, then $\tan(t_1)=\tan(t_2)$ as well.
However, on the interval $]-\pi/2,\pi/2[$, the tangent is a bijection, hence a self-intersection must occur at $x=0$. Therefore $\cos^2 t_1=\cos^2 t_2=\frac14$, and $t_{1,2}=\pm\frac{\pi}{3}$.
Your mistake comes from the conclusion that $\tan t_1=-\tan t_1$. You can't factor out $1-4\cos^2t_1$ if it's zero.