This is a quite elementary and simplified equation, but I am afraid that I am missing some of the solutions.
I have
$ t = t\cos\theta – n \sin\theta$.
What I had done before to solve it was the following,
$t^2 = t^2 \cos^2 \theta – 2tn\cos\theta\sin\theta + n^2\sin^2\theta\\$,
$t^2 \sin^2\theta = -2tn\cos\theta\sin\theta + n^2 \sin^2\theta$,
$(t^2-n^2)\sin^2\theta = -2tn\cos\theta\sin\theta$,
$(t^2-n^2)\tan\theta = -2tn$
In this very last step, I would just divide by $(t^2-n^2)$ and solve for $\theta$ it being the inverse tangent of what precedes in the equation. However, I am afraid that I am leaving out potential solutions where $(t^2-n^2) = 0$ or worse when $-2tn/(t^2-n^2)$ may be undefined (as $0/0$).
I know that from the initial equation, I can do the following,
$ t(1 – \cos\theta) = -n\sin\theta$.
Here I am also afraid of dividing by n or t, because they can individually be zero. In this instance where $n=0$,
I get that $t=0$, and $\cos\theta = 1$. But I am not sure if I am leaving out other possibilities.
Best Answer
The first problem was dividing by $\sin\theta\cos\theta,$ where those could be zero. So handle the cases when $\sin\theta=0$ and $\cos\theta=0$ first.
From:$$(t^2-n^2)\sin^2\theta =-2tn\cos\theta\sin\theta\tag1$$
If $\sin\theta=0,$ these are always equal.
But the earlier squaring added the case when $\sin\theta=0,\cos\theta=-1.$
When $\cos \theta=0,$ you need $t^2-n^2=0,$ which means $n=\pm t.$ Again, squaring both sides added cases, here adding the case where $\sin \theta =\pm 1,$ $n=t\sin\theta.$
When $\sin\theta\cos\theta\neq 0,$ you can divide by it, and then $\tan\theta$ is defined and not equal to $0.$
If $t^2-n^2=0,$ this equation has a solution only when $n=t=0.$ Then any $\theta$ works.
The easier solution is to never square and convert the equation to:
$$\tan(\theta/2)=\frac{1-\cos\theta}{\sin\theta}=-\frac nt$$ Handle the cases where $\sin(\theta)=0$ and $t=0$ seperately.