You're looking for [tex]t_1\neq t_2[/tex] such that [tex](x(t_1),y(t_1))=(x(t_2),y(t_2))[/tex].
[tex]\begin{cases}2\cos t_1=2\cos t_2\\\sin t_1-\sin2t_1=\sin t_2-\sin2t_2\end{cases}[/tex]
Recall that [tex]\sin2x=2\sin x\cos x[/tex], so the second equation can be written as
[tex]\sin t_1-2\sin t_1\cos t_1=\sin t_2-2\sin t_2\cos t_2[/tex]
[tex]\sin t_1(1-2\cos t_1)=\sint t_2(1-2\cos t_2)[/tex]
Since [tex]2\cos t_1=2\cos 2_t[/tex], and assuming [tex]1-2\cos t_2\neq0[/tex], you get
[tex]\sin t_1(1-2\cos t_1)=\sint t_2(1-2\cos t_1)[/tex]
[tex](\sin t_1-\sin t_2)(1-2\cos t_1)=0[/tex]
which admits two possibilities; either [tex]\sin t_1=\sin t_2[/tex] or [tex]1-2\cos t_1=0[/tex]. In the first case, since we're assuming [tex]t_1\neq t_2[/tex], we can use the fact that [tex]\sin(\pi-x)=\sin x[/tex] to arrive at a solution of [tex]t_2=\pi-t_1[/tex].
In the second case, you have
[tex]1-2\cos t_1=0\implies \cos t_1=\dfrac12\implies t_1=\dfrac\pi3\text{ or }\dfrac{5\pi}3[/tex]
Let's check which of these solutions work. If [tex]t_1=\dfrac\pi3[/tex], then the sine equation suggests [tex]t_2=\pi-\dfrac\pi3=\dfrac{2\pi}3[/tex]. However,
[tex]\left(x\left(\dfrac\pi3\right),y\left(\dfrac\pi3\right)\right)=(1,0)[/tex]
[tex]\left(x\left(\dfrac{2\pi}3\right),y\left(\dfrac{2\pi}3\right)\right)=(-1,\sqrt3)[/tex]
so in fact this is an extraneous solution. So let's return to the first equation in the system,
[tex]2\cos t_1=2\cos t_2\implies \cos t_1=\cos t_2[/tex]
Again, assuming [tex]t_1\neq t_2[/tex], we can use the fact that [tex]\cos(2\pi-x)=\cos x[/tex] to arrive at a solution of [tex]t_2=2\pi-t_2[/tex]. Now, if [tex]t_1=\dfrac\pi3[/tex], we get [tex]t_2=\dfrac{5\pi}3[/tex]. Let's check if this works:
[tex]\left(x\left(\dfrac\pi3\right),y\left(\dfrac\pi3\right)\right)=(1,0)[/tex]
[tex]\left(x\left(\dfrac{5\pi}3\right),y\left(\dfrac{5\pi}3\right)\right)=(1,0)[/tex]
Indeed, this solution works! So the curve intersects itself at the point (1,0), which the curve passes for the first time through when [tex]t=\dfrac\pi3[/tex] and the second time when [tex]t=\dfrac{5\pi}3[/tex].
Now, to find the tangent line, we need to compute the derivative of [tex]y[/tex] with respect to [tex]x[/tex]. You have
[tex]\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}[/tex]
When [tex]t=\dfrac\pi3[/tex], you have a slope of [tex]-\dfrac{\sqrt3}2[/tex]; at [tex]t=\dfrac{5\pi}3[/tex], the slope is [tex]\dfrac{\sqrt3}2[/tex].
The tangent lines are then
[tex]y_1-0=-\dfrac{\sqrt3}2(x-1)\implies y_1=-\dfrac{\sqrt3}2x+\dfrac{\sqrt3}2[/tex]
[tex]y_2-0=\dfrac{\sqrt3}2(x-1)\implies y_2=\dfrac{\sqrt3}2x-\dfrac{\sqrt3}2[/tex]
