Let P be the image of the point (–3, 2) with respect to x-axis. Keeping the origin as same, the coordinate axes are rotated through an angle 60° in the clockwise sense. The coordinates of point P with respect to the new axes are

Since P is the image of the point (–3, 2) with respect to x-axis, therefore, the coordinates of P are (–3, –2).
Let (x′, y′ ) be the coordinates of P with respect to new axes. Then,
$x\text{'}=x \cos \alpha +y \sin \alpha =-3\cos (-{60}^0)-2 \sin (-{60}^0)$

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

$y\text{'}=x \sin \alpha +y \cos \alpha =3 \sin (-{60}^0)-2 \cos (-{60}^0)$

$=\left(\frac{3\sqrt{3}+2}{2}\right)$

$\therefore$ Coordinates of P are $\left[\frac{2\sqrt{3}-3}{2},-\left(\frac{3\sqrt{3}+2}{2}\right)\right]$