I am kind of stuck with the solving process of this question:
\begin{align}
& \text{error } = 0.01 \\
\implies & \tan\theta = \text{ error }+ \sin\theta \\
\implies & \tan\theta = 0.01 + \sin\theta \\
\implies & \tan\theta - \sin\theta = 0.01 \\
\implies & \dfrac {\sin\theta}{\cos\theta} - \sin\theta = 0.01
\end{align}
And this is where I stop, I don't know where to go from here.
On
The greatest value of $x$ is $\tan^{-1} 0.275$ radians, or something infinitely smaller than that.
Plotting the graph $round(100 \tan x) - round(100 \sin x)$ on Desmos gives many solutions, but the greatest solution that satisfies $x < \pi/2$ is approximately $0.26386$.
However, $\tan 0.26386$ $\approx 0.2749933$, and $\sin 0.26386 \approx 0.26515$. This suggests that $\tan x$ can get as close to $0.275$ as you want, and $\sin x$ will still round to $0.27$. However, once $\tan x$ is $0.275$, then it rounds to $0.28$, but $\sin x$ still rounds to $0.27$.
The idea of this formula is to round $\tan x$ and $\sin x$ to $2$ significant figures, and find $x$ where they are equal, or in other words, find the zeroes of $round(\tan x) - round(\sin x)$.
Since in Desmos, the function $round()$ rounds only to the nearest integer, then we can multiply by $100$, round it, and then divide back by $100$, which results in $\frac{1}{100}(100 \ round(\tan x)$ $- 100\ round(\tan x)) = 0$, and simplifies to the formula above.
On
Robert Israel gave the rigorous answer; the solution implies solving a quartic. Zwin gave also an approximation.
Just for the fun of it, let me give another one.
Assuming that $r$ and $x$ are small, we could use Taylor series giving $$\tan(x)-\sin(x)=\frac{x^3}{2}+O\left(x^5\right)$$ then $$\tan(x)-\sin(x)=r\implies \frac{x^3}{2}=r\implies x=\sqrt[3]{2r}\tag 1$$ which exactly Zwin's result.
We can do a little better, using, instead of Taylor series, Padé approximants and get $$\tan(x)-\sin(x) \approx \frac{2 x^3}{4-x^2}$$ You will be amazed to see how close rhs and lhs are if you plot the two functions for $-1 \leq x \leq 1$. So, using it $$\tan(x)-\sin(x)=r\implies \frac{2 x^3}{4-x^2}=r\implies 2x^3+ rx^2-4 r=0$$
So, using approximations, we reduced the problem to a cubic equation which has the good idea to show only one real root. Using this, we get as an explicit solution $$x=\frac{1}{3} r \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{216}{r^2}-1\right)\right)\tag 2$$
The pitty is that for your case, $(1)$ and $(2)$ lead to almost identical results.
Edit
By analogy with the result from $(1)$, let us assume that we can write $$x=\sum_{n=1}^p a_n\left(\sqrt[3]{2r}\right)^n$$ and use Taylor series for the expansion of $\tan(x)-\sin(x)=r$ around $r=0$. This leads to $$a_1=1 \qquad a_3=-\frac{1}{12}\qquad a_5=-\frac{1}{120}\qquad a_7=\frac{1}{336}\qquad a_9=\frac{1}{2304}$$ all even coefficients being $0$.
As shown in the table below, the results are extremly good even for "large" values of $r$. $$\left( \begin{array}{ccc} r& \text{approximation} & \text{exact} \\ 0.005 & 0.214606 & 0.214606 \\ 0.010 & 0.269763 & 0.269763 \\ 0.015 & 0.308200 & 0.308200 \\ 0.020 & 0.338625 & 0.338625 \\ 0.025 & 0.364183 & 0.364183 \\ 0.030 & 0.386414 & 0.386414 \\ 0.035 & 0.406202 & 0.406202 \\ 0.040 & 0.424105 & 0.424105 \\ 0.045 & 0.440501 & 0.440501 \\ 0.050 & 0.455660 & 0.455660 \\ 0.055 & 0.469783 & 0.469783 \\ 0.060 & 0.483021 & 0.483021 \\ 0.065 & 0.495495 & 0.495495 \\ 0.070 & 0.507300 & 0.507300 \\ 0.075 & 0.518513 & 0.518513 \\ 0.080 & 0.529200 & 0.529200 \\ 0.085 & 0.539414 & 0.539414 \\ 0.090 & 0.549200 & 0.549200 \\ 0.095 & 0.558598 & 0.558597 \\ 0.100 & 0.567640 & 0.567639 \\ 0.105 & 0.576356 & 0.576355 \\ 0.110 & 0.584771 & 0.584770 \\ 0.115 & 0.592908 & 0.592907 \\ 0.120 & 0.600787 & 0.600786 \\ 0.125 & 0.608424 & 0.608423 \\ 0.130 & 0.615837 & 0.615836 \\ 0.135 & 0.623039 & 0.623038 \\ 0.140 & 0.630043 & 0.630041 \\ 0.145 & 0.636861 & 0.636859 \\ 0.150 & 0.643504 & 0.643501 \\ 0.155 & 0.649980 & 0.649977 \\ 0.160 & 0.656299 & 0.656296 \\ 0.165 & 0.662469 & 0.662465 \\ 0.170 & 0.668497 & 0.668493 \\ 0.175 & 0.674390 & 0.674386 \\ 0.180 & 0.680155 & 0.680150 \\ 0.185 & 0.685797 & 0.685792 \\ 0.190 & 0.691323 & 0.691317 \\ 0.195 & 0.696736 & 0.696729 \\ 0.200 & 0.702042 & 0.702035 \end{array} \right)$$
On
Using zwin's suggestion, using $t=\tan \left(\frac{x}{2}\right)$, $$\tan(x)-\sin(x)-r=0\implies r t^4+4 t^3-r=0$$ We can solve the quartic equation with radicals (ugly/monstreous solutions) and develop the valid one as Taylor series around $r=0$. Using $z=\sqrt[3]{2r}$, this leads to $$t=\frac{z}{2}-\frac{z^5}{96}+\frac{z^9}{1536}-\frac{35 z^{13}}{663552}+\frac{77 z^{17}}{15925248}+O\left(z^{21}\right)$$
Now, using $x=2 \tan ^{-1}(t)$ and repeating the expansion, we get
$$x=z-\frac{z^3}{12}-\frac{z^5}{120}+\frac{z^7}{336}+\frac{z^9}{2304}-\frac{5 z^{11}}{25344}-\frac{35 z^{13}}{1078272}+\frac{z^{15}}{61440}+\frac{385 z^{17}}{135364608}+O\left(z^{19}\right)$$
For $r=\frac 12$ that is to say $z=1$, the above expansion gives as an estimate $x=\frac{617563433269}{677499863040}\approx 0.911533$ while the "exact" solution would be $0.911531$.
For $r=\frac 1{100}$, this would lead to the approximation $$x=0.269763141387359251$$ while the "exact" solution would be $$x=0.269763141387359224$$
It $\sin(\theta) = x \in (0,\pi/2)$, $\tan(\theta) = x/\sqrt{1-x^2}$. The equation $\tan(\theta) - \sin(\theta) = r$ then implies $${x}^{4}+2\,r{x}^{3}+{r}^{2}{x}^{2}-2\,rx-{r}^{2} = 0$$ For $r = 1/100$ this is irreducible. It does have closed-form solutions in terms of radicals, but they are not pleasant. Numerical methods can be used.