You're looking for \(t_1
eq t_2\) such that \((x(t_1),y(t_1))=(x(t_2),y(t_2))\).
\(\begin{cases}2\cos t_1=2\cos t_2\\\sin t_1-\sin2t_1=\sin t_2-\sin2t_2\end{cases}\)
Recall that \(\sin2x=2\sin x\cos x\), so the second equation can be written as
\(\sin t_1-2\sin t_1\cos t_1=\sin t_2-2\sin t_2\cos t_2\)
\(\sin t_1(1-2\cos t_1)=\sint t_2(1-2\cos t_2)\)
Since \(2\cos t_1=2\cos 2_t\), and assuming \(1-2\cos t_2
eq0\), you get
\(\sin t_1(1-2\cos t_1)=\sint t_2(1-2\cos t_1)\)
\((\sin t_1-\sin t_2)(1-2\cos t_1)=0\)
which admits two possibilities; either \(\sin t_1=\sin t_2\) or \(1-2\cos t_1=0\). In the first case, since we're assuming \(t_1
eq t_2\), we can use the fact that \(\sin(\pi-x)=\sin x\) to arrive at a solution of \(t_2=\pi-t_1\).
In the second case, you have
\(1-2\cos t_1=0\implies \cos t_1=\dfrac12\implies t_1=\dfrac\pi3\text{ or }\dfrac{5\pi}3\)
Let's check which of these solutions work. If \(t_1=\dfrac\pi3\), then the sine equation suggests \(t_2=\pi-\dfrac\pi3=\dfrac{2\pi}3\). However,
\(\left(x\left(\dfrac\pi3\right),y\left(\dfrac\pi3\right)\right)=(1,0)\)
\(\left(x\left(\dfrac{2\pi}3\right),y\left(\dfrac{2\pi}3\right)\right)=(-1,\sqrt3)\)
so in fact this is an extraneous solution. So let's return to the first equation in the system,
\(2\cos t_1=2\cos t_2\implies \cos t_1=\cos t_2\)
Again, assuming \(t_1
eq t_2\), we can use the fact that \(\cos(2\pi-x)=\cos x\) to arrive at a solution of \(t_2=2\pi-t_2\). Now, if \(t_1=\dfrac\pi3\), we get \(t_2=\dfrac{5\pi}3\). Let's check if this works:
\(\left(x\left(\dfrac\pi3\right),y\left(\dfrac\pi3\right)\right)=(1,0)\)
\(\left(x\left(\dfrac{5\pi}3\right),y\left(\dfrac{5\pi}3\right)\right)=(1,0)\)
Indeed, this solution works! So the curve intersects itself at the point (1,0), which the curve passes for the first time through when \(t=\dfrac\pi3\) and the second time when \(t=\dfrac{5\pi}3\).
Now, to find the tangent line, we need to compute the derivative of \(y\) with respect to \(x\). You have
\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{\cos t-2\cos2t}{-2\sin t}=\dfrac{2\cos2t-\cos t}{2\sin t}\)
When \(t=\dfrac\pi3\), you have a slope of \(-\dfrac{\sqrt3}2\); at \(t=\dfrac{5\pi}3\), the slope is \(\dfrac{\sqrt3}2\).
The tangent lines are then
\(y_1-0=-\dfrac{\sqrt3}2(x-1)\implies y_1=-\dfrac{\sqrt3}2x+\dfrac{\sqrt3}2\)
\(y_2-0=\dfrac{\sqrt3}2(x-1)\implies y_2=\dfrac{\sqrt3}2x-\dfrac{\sqrt3}2\)
