The equation of the normal to the curve x = sin t, y = cos 2t at $t=\pi /6 is$

$$\begin{gathered} x=\sin t, y=\cos 2t, t=\frac{\mathrm{\pi }}{6} \\ P\left(\frac{1}{2}, \frac{1}{2}\right) \\ \frac{dx}{dt}=\cos t \\ \frac{dy}{dt}=-2\sin 2t \\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{2\sin 2t}{\cos t}=-2 \\ equation of normal y-y_1=m^1(x-x_1) \\ y-\frac{1}{2}=\frac{1}{2}\left(x-\frac{1}{2}\right) \\ 2y-1=\frac{2x-1}{2} \\ 2x-4y+1=0 \end{gathered}$$