x = sin3t ⇒

dx

dt

= 3 cos 3t
∴ x(t=

π

4

) = sin 3 (

π

4

) =

1

√2

y = cos 2t
⇒

dy

dt

= - 2 sin 2t
∴ y(t=

π

4

) = cos 2t = cos 2 (

π

4

) = 0
⇒

dy

dx

=

dy

dt

.

dt

dx

= - 2 sin 2t

1

3cps3t

= -

2

3

(

sin2t

cos3t

)
∴

dy

dx(t= π 4 )

= −

2

3

sin(2×π4)

cos(3×π4)

= −

2

3

sin π 2

cos 3π 4

= -

2

3

|

1

− 1 √2

| =

2√2

3

Therefore, the equation of the tangent at the point (

1

√2

,0) is
y - 0 =

2√2

3

(x−

1

√2

)
y =

2√2

3

x−

2

3

3y - 2 √2x + 2 = 0