Find the equation of tangent to the curve `x=sin3t , y=cos2t\ \ ` at `t=pi/4`

Explanation

`x=sin(3/4x),y=cos(2/4x)`

`x=sin(x-x/4),y=cos(x/2)`

`x=sin(x/4)=1/sqrt2,y=0`

Slope of target=`dy/dx)_(x_o,y_o)=((dy/dt)/(dx/dt))`

`dy/dt=d/dt(cost)`

`=-sin2t*2`

`(dy/dx)_(t=x/4)=(-2sin2t)`

`(dy/dx)_(1/sqrt2,0)=-2/(-3/sqrt2)=(2sqrt2)/3`

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