range of $3x^2-2xy$ subjected to $x^2+y^2=1$ If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done  try here is  use arithmetic geometric inequality
$\displaystyle x^2+y^2\geq 2xy$
$\displaystyle -2xy\geq -(x^2+y^2)$
$\displaystyle 3x^2-2xy\geq 2x^2-y^2$
this will not help more
how do i solve it help  me please

For a calculs-free approach, note that
$$
\frac{3+\sqrt{13}}{2}\left[\frac{2y-(3+\sqrt{13})x}{\sqrt{26+6\sqrt{13}}}\right]^2+\frac{3-\sqrt{13}}{2}\left[\frac{2y-(3-\sqrt{13})x}{\sqrt{26-6\sqrt{13}}}\right]^2=3x^2-2xy
$$
and
$$
\left[\frac{2y-(3+\sqrt{13})x}{\sqrt{26+6\sqrt{13}}}\right]^2+\left[\frac{2y-(3-\sqrt{13})x}{\sqrt{26-6\sqrt{13}}}\right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $\dfrac{3\pm\sqrt{13}}{2}$.
