Question

The curve x = 2 cos(t), y = sin(t) − sin(2t), 0 ≤ t ≤ 2π crosses itself once,find this intersection point and then find the equations of the two tangent lines at the intersection point.

Answer 1

You're looking for $t_1\neq t_2$ 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 $1-2\cos t_2\neq0$, 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 $t_1\neq t_2$, 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 $t_1\neq t_2$, 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

$y_1-0=-\dfrac{\sqrt3}2(x-1)\implies y_1=-\dfrac{\sqrt3}2x+\dfrac{\sqrt3}2$
$y_2-0=\dfrac{\sqrt3}2(x-1)\implies y_2=\dfrac{\sqrt3}2x-\dfrac{\sqrt3}2$