x = sin3t ⇒ $\frac{dx}{dt}$ = 3 cos 3t ∴ $x_{t=\pi/4}$ = sin 3 $\left(\frac{\pi}{4}\right)$ = $\frac{1}{\sqrt{2}}$ y = cos 2t ⇒ $\frac{dy}{dt}$ = - 2 sin 2t ∴ $y_{t=\pi/4}$ = cos 2t = cos 2 $\left(\frac{\pi}{4}\right)$ = 0 ⇒ $\frac{dy}{dx}$ = $\frac{dy}{dt} \cdot \frac{dt}{dx}$ = - 2 sin 2t $\frac{1}{3\text{cps}3t}$ = - $\frac{2}{3}\left(\frac{\sin 2t}{\cos 3t}\right)$ ∴ $\frac{dy}{dx}_{t=\pi/4}$ = $-\frac{2}{3} \frac{\sin(2 \times \pi 4)}{\cos(3 \times \pi 4)}$ = $-\frac{2}{3} \frac{\sin\frac{\pi}{2}}{\cos\frac{3\pi}{4}}$ = - $\frac{2}{3} \left| \frac{1}{-\frac{1}{\sqrt{2}}} \right|$ = $\frac{2\sqrt{2}}{3}$ Therefore, the equation of the tangent at the point $\left(\frac{1}{\sqrt{2}}, 0\right)$ is y - 0 = $\frac{2\sqrt{2}}{3} \left( x - \frac{1}{\sqrt{2}} \right)$ y = $\frac{2\sqrt{2}}{3}x - \frac{2}{3}$ 3y - 2 $\sqrt{2x}$ + 2 = 0