$x=t-\sin (t)$. How do I solve for $t$?

Question

I'm working on eliminating the parameter for $$ \begin{pmatrix} t\cos \frac{3\pi}{4}-\sin t\sin \frac{3\pi}{4} \\ t\sin \frac{3\pi}{4}+\sin t\cos \frac{3\pi}{4} \end{pmatrix}$$

I'm getting stuck at solving for t as a function of $x$, specifically at $x\sqrt{2}=t-\sin \left(t\right)$ where I just can't find a way to do it. I don't really want to use a Taylor polynomial to solve simply because I haven't taken Algebra 2 Trig yet (I'm a freshman) and solving those polynomials by hand would be very tedious.

I've also tried logarithms to get rid of the subtraction,

$x\sqrt{2}=t-\sin \left(t\right)$

$x\sqrt{2}=\ln \left(e^t\right)-\ln \left(e^{\sin \left(t\right)}\right)$

$x\sqrt{2}=\ln \left(\frac{e^t}{e^{\sin \left(t\right)}}\right)$

$e^{x\sqrt{2}}=\frac{e^t}{e^{\sin \left(t\right)}}$

$e^{x\sqrt{2}}=e^{t-\sin \left(t\right)}$

And then I'm no better off than I started.

I was working on another problem (solving $n \cdot x^{n-1}>e^x$) and I heard I needed the Product Log function and so if I need some kind of analytic function like that then I can figure that out. I just can't see a way around the t being added. I also think I read somewhere that it can't be expressed as a finite series of elementary functions so it might just have to be Taylor polynomials but if anyone has any insight that would be much appreciated.

Answer 1

$x$ should read:

$$x=-\frac{t+\sin t}{\sqrt{2}}$$

Since you've mentioned eliminating $t$ and assume you're not tackling an advanced problem. I suspect that you're required to find an implicit equation for $x$ and $y$.

Now

\begin{align} y &= \frac{t-\sin t}{\sqrt{2}} \\ x+y &= -\sqrt{2} \sin t \\ y-x &= \sqrt{2} t \\ \end{align}

Therefore

$$\fbox{$\sin \frac{x-y}{\sqrt{2}}=\frac{x+y}{\sqrt{2}}$}$$

which is a rotation of a sine curve by $45^{\circ}$.

Answer 2

EDIT1:

Your question is incomplete. Whereas you defined one curve's $(x,y)$, the $y$ parametrization of the second curve is totally missing. So no solution.

I supplied an arbitrary $y$ as below. It may be noted that some points of real intersection now exist.Red first curve is the one you gave, the blue second one $y$ part I assumed.

$$ x = \frac{-(t+\sin t)}{\sqrt{2}};\, y = \frac{(t-\sin t)}{\sqrt{2}} $$

$$ x = \frac{-(t+\sin t)}{\sqrt{2}}; \, y = {(t+\cos t)} $$