I need to find the self-intersection of the curve
$$ C_1:x=t^2-5t+4, y=4\sin({\pi t\over 2}), 0 \le t \le 6 $$
I figured I would try to solve these equations on my TI-nspire:
$$t_1^2-5t_1+4=t_2^2+5t_2+4$$
$$4\sin({\pi t_1 \over 2})=4\sin({\pi t_2 \over 2})$$
for $t_1$ and $t_2$, but it gives me $t_1=-{(4n-13) \over 2}$ and $t_2={4n-3 \over 2}$ and I'm not sure how to proceed from here? why do I see an $n$?
Thanks!
From intersection on the $x$ coordinate (you have a typo in your work) $${\it t_1}^2-5\,{\it t_1}+4={\it t_2}^2-5\,{\it t_2}+4$$ gives $$(t_1-t_2)(t_1+t_2-5)$$
So we want $$t_1 = 5-t_2$$
From the $y$ coordinates we want $$\sin \left({{\pi\,{t_1}}\over{2}}\right)=\sin \left({{\pi\, { t_2}}\over{2}}\right)$$ So either $$ {{\pi\,{t_1}}\over{2}} + {{\pi\,{t_2}}\over{2}} = \pi ~~ or \\ {{\pi\,{t_1}}\over{2}} =2 \pi - {{\pi\,{t_2}}\over{2}} $$ First is not possible since $t_1+t_2=5$. Hence only the second holds. So we have $$ t_1 = 4 + t_2, ~~~\text{and} ~~~t_1 + t_2 = 5$$ which gives $$ { t_1}={{9}\over{2}} , { t_2}={{1}\over{2}} $$ and the point of intersection is $$\left( {{7}\over{4}} , {{4}\over{\sqrt{2}}} \right) $$