Find the equation of the tangent to the curve

finding the slope of the tangent by differentiating the curve

$\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}$

equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$

now comparing the slope of a tangent with the given equation

$\mathrm{m}($ tangent $)=2$

$\frac{3}{2 \sqrt{3 x-2}}=2$

$\frac{9}{16}=3 x-2$

$x=\frac{41}{48}$

since this point lies on the curve, we can find $y$ by substituting $x$

$y=\sqrt{\frac{41}{16}-2}$

$y=\frac{3}{4}$

therefore, the equation of the tangent is

$y-\frac{3}{4}=2\left(x-\frac{41}{48}\right)$