x = sin3t ⇒ $\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}$ = 3 cos 3t
∴ ${\mathit{x}}_{\left(\mathit{t}=\frac{\mathit{\pi }}{4}\right)}$ = sin 3 $\left(\frac{\mathit{\pi }}{4}\right)$ = $\frac{1}{\sqrt{2}}$
y = cos 2t
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}$ = - 2 sin 2t
∴ ${\mathit{y}}_{\left(\mathit{t}=\frac{\mathit{\pi }}{4}\right)}$ = cos 2t = cos 2 $\left(\frac{\mathit{\pi }}{4}\right)$ = 0
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}.\frac{\mathit{d}\mathit{t}}{\mathit{d}\mathit{x}}$
= - 2 sin 2t $\frac{1}{3\mathit{c}\mathit{p}\mathit{s}3\mathit{t}}$
= - $\frac{2}{3}\left(\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{t}}{\mathit{c}\mathit{o}\mathit{s}3\mathit{t}}\right)$
∴ $\frac{\mathit{d}\mathit{y}}{{\mathit{d}\mathit{x}}_{\left(\mathit{t}=\frac{\mathit{\pi }}{4}\right)}}$ = $-\frac{2}{3}\frac{\mathit{s}\mathit{i}\mathit{n}\left(2\times \mathit{\pi }4\right)}{\mathit{c}\mathit{o}\mathit{s}\left(3\times \mathit{\pi }4\right)}$
= $-\frac{2}{3}\frac{\mathit{s}\mathit{i}\mathit{n}\frac{\mathit{\pi }}{2}}{\mathit{c}\mathit{o}\mathit{s}\frac{3\mathit{\pi }}{4}}$
= - $\frac{2}{3}|\frac{1}{-\frac{1}{\sqrt{2}}}|$ = $\frac{2\sqrt{2}}{3}$
Therefore, the equation of the tangent at the point $\left(\frac{1}{\sqrt{2}},0\right)$ is
y - 0 = $\frac{2\sqrt{2}}{3}\left(\mathit{x}-\frac{1}{\sqrt{2}}\right)$
y = $\frac{2\sqrt{2}}{3}\mathit{x}-\frac{2}{3}$
3y - 2 $\sqrt{2\mathit{x}}$ + 2 = 0