Eliminate $\theta$ from $4x=5\cos\theta -\cos 5\theta$ and $4y=5\sin\theta -\sin 5\theta$ Eliminate $\theta$ from the equations.
$$4x=5\cos\theta -\cos 5\theta$$
$$4y=5\sin\theta -\sin 5\theta$$
Alternative forms are
$$x=5\cos^3\theta-4\cos^5\theta$$
$$y=5\sin^3\theta-4\sin^5\theta$$
and $$x=\cos^3\theta(3-2\cos2\theta)$$
$$y=\sin^3\theta(3+2\cos2\theta)$$

Let $c=\cos2\theta$.
$$4x=5\cos\theta -\cos 5\theta\tag1$$
$$4y=5\sin\theta -\sin 5\theta\tag2$$
Squaring and adding ,$$16(x^2+y^2)=25+1-10\cos4\theta$$ $$8(x^2+y^2)=13-5(2c^2-1)$$ $$5c^2=9-4(x^2+y^2)=k$$
Squaring and subtracting, $$16(x^2-y^2)=25\cos2\theta+\cos10\theta-10\cos6\theta$$ $$16(x^2-y^2)=25c+(16c^5-20c^3+5c)-10(4c^3-3c)$$ $$4(x^2-y^2)=c(4c^4-15c^2+15)$$
Substituting $k$,
$$100(x^2-y^2)=c(4k^2-75k+375)$$
Further simplifying leads to, $$25(x^2-y^2)=c(16\lambda^2+3\lambda+6)$$ where $\lambda=(x^2+y^2)$.
Squaring again and replacing all values finally we get, $$3125(x^2-y^2)^2=[9-4(x^2+y^2)][16(x^2+y^2)^2+3(x^2+y^2)+6]^2$$
And this is that epicycloid, as being described in @orangeskid's answer!
