Find the number of solutions of the equation $G=G'$ in $[0,2\pi]$ Let $G$ be the sum of infinite geometric series whose first term is $\sin\theta$ and common ratio is $\cos\theta$,while $G'$ be the sum of a different infinite geometric series whose first term is $\cos\theta$ and common ratio is $\sin\theta$.Find the number of solutions of the equation $G=G'$ in $[0,2\pi]$

I found $G=\frac{\sin\theta}{1-\cos\theta}$ and $G'=\frac{\cos\theta}{1-\sin\theta}$
According to the question,
$\frac{\sin\theta}{1-\cos\theta}=\frac{\cos\theta}{1-\sin\theta}$
$\sin\theta-\sin^2\theta=\cos\theta-\cos^2\theta$
$(\sin\theta-\cos\theta)(1-\sin\theta-\cos\theta)=0$
Either $\sin\theta-\cos\theta=0$ or $\sin\theta+\cos\theta=1$
$\sin\theta+\cos\theta=1$ gives $\theta=0,\frac{\pi}{2},2\pi$ but these value are not allowed as they make denominators zero.
$\sin\theta-\cos\theta=0$ gives $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.So there are two solutions.
But the book gives answer as no solutions.I do not understand why?Please help me point my mistake.

So we have $G=\frac{\sin\theta}{1-\cos\theta}$ and $G'=\frac{\cos\theta}{1-\sin\theta}$,
an important condition for the sum of the geometric series is $-1<r<1$ where r is the common ratio, thus
$$\frac{\sin\theta}{1-\cos\theta}=\frac{\cos\theta}{1-\sin\theta}$$
$${\sin\theta}*(1-\sin\theta)={\cos\theta}*(1-\cos\theta)$$
$${\sin\theta}-{\cos\theta}=({\sin\theta})^2-({\cos\theta})^2$$
$${\sin\theta}-{\cos\theta}=({\sin\theta}-{\cos\theta})*({\sin\theta}+{\cos\theta})$$
$$1=({\sin\theta}+{\cos\theta})$$ with $({\sin\theta}-{\cos\theta})$ not zero.
Now we see that this equation holds for $\theta=\pi/2$ or $\theta=0$, but those solutions are in contradiction with initial conditions about the common ratio.
The other possible solution is $({\sin\theta}-{\cos\theta})=0$ thus $\theta=\pi/4$. I don't see contradictions with the initial conditions here, so this is the solution.
If we substitute the solution in the first equation we get
$$1+\sqrt{2}=1+\sqrt{2}$$
