How we solve $0=y\sin\frac{3t}{2}+x\cos\frac{3t}{2}-y\sin(t)-x\cos(t)$ for real $t$ ($\frac{4\pi}{3}I have a system of equation:

*

*$x^2+y^2=y\sin\frac{3t}{2}+x\cos\frac{3t}{2}$

*$x^2+y^2=y\sin(t)+x\cos(t)$
I want to get an algebraic curve depending only from $x$ and $y$. If it is too complex, it would be already sufficient to determine, which degree the resulting curve has.
Can we at least determine the degree (the largest power $n$ that appears in $x^n$) of the curve equation in $x$ and $y$?
I am interested only in a (real) range $\frac{4\pi}{3}<t<2\pi$, which hopefully gives a condition that reduces the effort in solving the equation.

You already isolated a linear equation in $x$ and $y$ for each value $t$:
$$ y \sin \frac{3t}{2} + x \cos \frac{3t}{2} - y \sin t - x \cos t = 0 $$
$$ y = \frac{\cos t - \cos \frac{3t}{2}}{\sin \frac{3t}{2} - \sin t} x $$
$$ y = m(t) \, x \quad\mathrm{\ where\ } m(t) = \frac{\cos t - \cos \frac{3t}{2}}{\sin \frac{3t}{2} - \sin t} $$
unless $\sin \frac{3t}{2} - \sin t = 0$, which occurs at $t=2\pi m$ and at $t=\frac{2\pi}{5} (2n+1)$ for any integer $m$ or $n$. This never happens in the given domain $\frac{4\pi}{3} < t < 2\pi$.
$$ m(t) = \frac{2 \sin \frac{5t}{4}\, \sin \frac{t}{4}}{2 \cos \frac{5t}{4}\, \sin \frac{t}{4}} $$
$$ m(t) = \tan \frac{5t}{4} $$
Based on this ratio, let's define $x(t)$ and $y(t)$ in terms of the unknown magnitude $r(t)$:
$$ x = r \cos \frac{5t}{4} $$
$$ y = r \sin \frac{5t}{4} $$
Then from the second original equation,
$$ r^2 = r \sin \frac{5t}{4}\, \sin t + r \cos \frac{5t}{4}\, \cos t $$
The solution $r=0$ is just the extra point $(0,0)$ which always satisfies the equations for any $t$. Ignoring that, we can divide out $r$.
$$ r = \cos \left(\frac{5t}{4} - t\right) = \cos \frac{t}{4} $$
So
$$ x = \cos \frac{5t}{4}\, \cos \frac{t}{4} $$
$$ x^2+y^2 = \cos^2 \frac{t}{4} $$
The fifth Chebyshev polynomial of the first kind is $T_5(z) = 16z^5 - 20z^3+5z$, so
$$ \cos \frac{5t}{4} = T_5\left(\cos \frac{t}{4}\right) = 16 \cos^5 \frac{t}{4} - 20 \cos^3 \frac{t}{4} + 5 \cos \frac{t}{4} $$
$$ x = 16 \cos^6 \frac{t}{4} - 20 \cos^4 \frac{t}{4} + 5 \cos^2 \frac{t}{4} $$
$$ x = 16 (x^2+y^2)^3 - 20 (x^2+y^2)^2 + 5 (x^2+y^2) $$
This is a polynomial equation in just $x$ and $y$ of degree 6.
