Step 1: Replace sine with cosine.
Use $\sin^2x=1-\cos^2x$ and let $t=\cos^2x$. This turns the equation into a quadratic in $t$.
Step 2: Substitute into the equation.
The equation $6\sin^2x=3\cos^4x-\sin^2x\cos^2x$ becomes $6(1-t)=3t^2-(1-t)t$.
Step 3: Expand both sides.
Left is $6-6t$. Right is $3t^2-t+t^2=4t^2-t$. So $6-6t=4t^2-t$.
Step 4: Form the quadratic.
Rearranged: $4t^2+5t-6=0$. Factor as $(4t-3)(t+2)=0$.
Step 5: Keep the valid root.
So $t=\dfrac34$ or $t=-2$. Since $t=\cos^2x$ cannot be negative, take $t=\dfrac34$, giving $\cos x=\pm\dfrac{\sqrt3}{2}$.
Step 6: Write the general solution.
$\cos x=\pm\dfrac{\sqrt3}{2}$ means $x=n\pi\pm\dfrac{\pi}{6}$ for any integer $n$. \[ \boxed{x=n\pi\pm\frac{\pi}{6}} \]