equations involving trig functions Is there a way to solve them by hand?
E.g: $2x^2 + 18y^2 - 4 + \sin(x^2+9y^2) = 0$
I can think of a way to express  $x^2 + 9y^2$ as $t^2$ but i can't proceed.
 A: If we let $x^2+9y^2=t$, we have
$$2t-4+\sin(t)=0$$
One can prove this has only one solution for $t$ around $1.5$, but we cannot solve for it algebraically.  Instead, using Newton's method with $t_0=1.5$, we get
$$t_{n+1}=t_n-\frac{2t_n-4+\sin(t_n)}{2+\cos(t_n)}$$
$$\begin{align}t_0&=1.5\\t_1&=1.501209721\\t_2&=1.501210073\\t_3&=1.501210073\end{align}$$
Thus, the solution is $t=1.501210073$ out that many decimals.  Then we are left with
$$x^2+9y^2=1.501210073$$
$$y=\frac13\sqrt{1.501210073-x^2}$$
and so the $(x,y)$ solutions are given by
$$(x,y)=\left(a,\frac13\sqrt{1.501210073-a^2}\right)$$

Notice that if $t$ were a rational multiple of $\pi$, we'd get a transcendental number on the LHS, but $0$ is algebraic, hence contradiction.  Particularly, $\sin(t)$ is algebraic for any rational multiple of $\pi$ due to the sum of angles formulas.
Thus, $t$ is not a rational multiple of $\pi$, but sadly if $t$ is not a rational multiple of $\pi$, the closed form for $\sin(t)$ is unknown, hence, we cannot "algebraically" solve the problem.
