\(x=r(\theta)\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r\sin\theta\)
\(y=r(\theta)\sin\theta\implies\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r\cos\theta\)
Since \(r=1-\cos\theta\), you have \(\dfrac{\mathrm dr}{\mathrm d\theta}=\sin\theta\). Now,
\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}=\dfrac{\sin^2\theta+(1-\cos\theta)\cos\theta}{\sin\theta\cos\theta-(1-\cos\theta)\sin\theta}\)
\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\cos\theta-\cos2\theta}{\sin2\theta-1}\)
When \(t=\dfrac\pi3\), you have
\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\cos\frac\pi3-\cos\frac{2\pi}3}{\sin\frac{2\pi}3-1}=-2\sqrt3-4\)
which will be the slope of the tangent line. At this point, you have \(r=1-\cos\dfrac\pi3=\dfrac12\), so now you need to find the corresponding x and y coordinates.
\(x=r\cos\theta=\dfrac12\cos\dfrac\pi3=\dfrac14\)
\(y=r\sin\theta=\dfrac12\sin\dfrac\pi3=\dfrac{\sqrt3}4\)
So the equation of the tangent line is
\(y-\dfrac{\sqrt3}4=(-2\sqrt3-4)\left(x-\dfrac14\right)\)
\(y=-2(2+\sqrt3)x+1+\dfrac{3\sqrt3}4\)
