How to find $y'(x=1)$ in parametric equation?

Question

Given parametric equation \begin{align} x(t)&=a-b(2t+\sin(2t))\\ y(t)&=b(1+\cos(2t)) \end{align} and $y(x=0)=1$.

Find $y'(x=1)$.

This is my effort.

\begin{align} \dfrac{dy}{dx}&=\dfrac{\dfrac{dy(t)}{dt}}{\dfrac{dx(t)}{dt}}\\ &=\dfrac{-2b\sin(2t)}{-b(2+2\cos(2t))}\\ &=\dfrac{\sin(2t)}{1+\cos(2t)} \end{align}

Now I confused to find $y'$ in variable $x$, because the parametric equation cannot explicitly express $y$ in $x$. So I can't find $y'(1)$. Anyone can give me hint to solve this problem?