Core approach
And then used the solution of Trigonometric Equation $\tan(θ)=\tan(β)$…
That sounds like a good approach to me. So what you're saying is that you got to
$$
5\pi\cos\alpha = n\pi+\tfrac\pi2-5\pi\sin\alpha,\quad n\in\mathbb Z
$$
and then solved this for $\alpha$? How exactly?
Tangent half-angle approach
Personally I would use the tangent half-angle formulas to turn this trigonometric equation into a polynomial one:
$$
t:=\tan\frac\alpha2\quad \sin\alpha=\frac{2t}{1+t^2}\quad \cos\alpha=\frac{1-t^2}{1+t^2}\\
5\frac{1-t^2}{1+t^2}=n+\frac12-5\frac{2t}{1+t^2}\\
10-10t^2=2n+2nt^2+1+t^2-20t\\
$$
So what values of $n$ should you be considering? Let'suse the fact that $\sin\alpha\in[-1,1]$ and the same for $\cos\alpha$.
$$5[-1\ldots 1]=n+\tfrac12-5[-1\ldots 1]\\n=5[-1\ldots 1]+5[-1\ldots 1]-\tfrac12$$
So a conservative estimate would be $n\in\{-10,-9,-8,\ldots,7,8,9\}$. Since you can't have both $\sin\alpha$ and $\cos\alpha$ be close to $\pm1$ at the same time, not all of these $n$ will have solutions, but this is good enough for now. Take each $n$ and compute the resulting $t$ (at most two for each $n$). You get $28$ different values.
$$
\begin{array}{rl|rr|r}
t && \alpha && n \\\hline
-18.88819 = & -\sqrt{79} - 10 & -3.035805 = & -173.93882° & -6 \\
-5.18925 = & -\frac{1}{3} \, \sqrt{31} - \frac{10}{3} & -2.760848 = & -158.18495° & -7 \\
-1.47741 = & \frac{1}{3} \, \sqrt{31} - \frac{10}{3} & -1.951541 = & -111.81505° & -7 \\
-1.11181 = & \sqrt{79} - 10 & -1.676584 = & -96.06118° & -6 \\
-0.90871 = & -\sqrt{119} + 10 & -1.475215 = & -84.52361° & -5 \\
-0.76274 = & -\frac{1}{3} \, \sqrt{151} + \frac{10}{3} & -1.303204 = & -74.66809° & -4 \\
-0.64575 = & -\sqrt{7} + 2 & -1.146765 = & -65.70481° & -3 \\
-0.54575 = & -\frac{1}{7} \, \sqrt{191} + \frac{10}{7} & -0.999154 = & -57.24732° & -2 \\
-0.45630 = & -\frac{1}{9} \, \sqrt{199} + \frac{10}{9} & -0.856168 = & -49.05481° & -1 \\
-0.37334 = & -\frac{1}{11} \, \sqrt{199} + \frac{10}{11} & -0.714628 = & -40.94519° & 0 \\
-0.29387 = & -\frac{1}{13} \, \sqrt{191} + \frac{10}{13} & -0.571642 = & -32.75268° & 1 \\
-0.21525 = & -\frac{1}{3} \, \sqrt{7} + \frac{2}{3} & -0.424031 = & -24.29519° & 2 \\
-0.13460 = & -\frac{1}{17} \, \sqrt{151} + \frac{10}{17} & -0.267592 = & -15.33191° & 3 \\
-0.04783 = & -\frac{1}{19} \, \sqrt{119} + \frac{10}{19} & -0.095581 = & -5.47639° & 4 \\
0.05294 = & -\frac{1}{21} \, \sqrt{79} + \frac{10}{21} & 0.105787 = & 6.06118° & 5 \\
0.19271 = & -\frac{1}{23} \, \sqrt{31} + \frac{10}{23} & 0.380745 = & 21.81505° & 6 \\
0.67686 = & \frac{1}{23} \, \sqrt{31} + \frac{10}{23} & 1.190052 = & 68.18495° & 6 \\
0.89944 = & \frac{1}{21} \, \sqrt{79} + \frac{10}{21} & 1.465009 = & 83.93882° & 5 \\
1.10046 = & \frac{1}{19} \, \sqrt{119} + \frac{10}{19} & 1.666377 = & 95.47639° & 4 \\
1.31107 = & \frac{1}{17} \, \sqrt{151} + \frac{10}{17} & 1.838389 = & 105.33191° & 3 \\
1.54858 = & \frac{1}{3} \, \sqrt{7} + \frac{2}{3} & 1.994827 = & 114.29519° & 2 \\
1.83233 = & \frac{1}{13} \, \sqrt{191} + \frac{10}{13} & 2.142438 = & 122.75268° & 1 \\
2.19152 = & \frac{1}{11} \, \sqrt{199} + \frac{10}{11} & 2.285425 = & 130.94519° & 0 \\
2.67853 = & \frac{1}{9} \, \sqrt{199} + \frac{10}{9} & 2.426964 = & 139.05481° & -1 \\
3.40290 = & \frac{1}{7} \, \sqrt{191} + \frac{10}{7} & 2.569951 = & 147.24732° & -2 \\
4.64575 = & \sqrt{7} + 2 & 2.717562 = & 155.70481° & -3 \\
7.42940 = & \frac{1}{3} \, \sqrt{151} + \frac{10}{3} & 2.874000 = & 164.66809° & -4 \\
20.90871 = & \sqrt{119} + 10 & 3.046012 = & 174.52361° & -5
\end{array}
$$
All of these look like valid solutions to me: they satisfy the initial equation. Since the tangent half-angle formulas can't represent $\alpha=\pi$ (it corresponds to $t=\infty$), we also need to check that this is not a solution. And of course these $\alpha$ are arguments to trigonometric functions, so adding any multiple of $2\pi$ will be a solution, too. The above are all the solutions in the $\alpha\in(-\pi,+\pi]$ range.
Trigonometric identities instead of tangent half-angle formulas
Update: After reading some other answers, and seeing how they avoid the tangent half-angle formulas, I wanted to look up the computation for that using well-established identities. Starting from the equation
\begin{align*}
5\pi\cos\alpha &= n\pi+\tfrac\pi2-5\pi\sin\alpha,\quad n\in\mathbb Z \\
\sin\alpha+\cos\alpha &= \frac{2n+1}{10}
\end{align*}
the sum on the left hand side is the most interesting part. Wikipedia list of trigonometric identities lists your $\tan\left(\tfrac\pi2-\theta\right)=\cot\theta$ under Reflections and also some formulas you can use to tackle that sum.
One approach uses shifts to turn $\cos$ into $\sin$ and product to sum identities in reverse to turn the sum into a product:
\begin{align*}
\cos\alpha &= \sin(\alpha+\tfrac\pi2) \\
\sin(\theta+\varphi)+\sin(\theta-\varphi)&=2\sin\theta\cos\varphi
\qquad\text{with }
\theta:=\alpha+\tfrac\pi4, \quad \varphi:=\tfrac\pi4 \\
\sin\alpha+\cos\alpha = \sin\alpha + \sin(\alpha+\tfrac\pi2) &= 2\sin(\alpha+\tfrac\pi4)\cos\tfrac\pi4 = \sqrt2\sin(\alpha+\tfrac\pi4)
\end{align*}
You might also start from a formula for angle sums:
\begin{align*}
\sin\alpha\cos\beta + \cos\alpha\sin\beta &= \sin(\alpha+\beta) \\
\beta := \tfrac\pi4 \qquad & \cos\beta=\sin\beta=\tfrac1{\sqrt2} \\
\tfrac1{\sqrt2}\left(\sin\alpha+\cos\alpha\right) &= \sin\left(\alpha+\tfrac\pi4\right)
\end{align*}
Either way you get
$$
\sin\alpha+\cos\alpha = \sqrt2\sin(\alpha+\tfrac\pi4) = \frac{2n+1}{10} \\
\sin(\alpha+\tfrac\pi4) = \frac{2n+1}{10\sqrt2} \\
\alpha = \arcsin\frac{2n+1}{10\sqrt2}-\frac14\pi
\qquad\text{or}\qquad
\alpha = \frac34\pi-\arcsin\frac{2n+1}{10\sqrt2}
\qquad\pmod{2\pi}
$$
where the second solution accounts for the fact that $\arcsin$ should be considered a multi-valued function, and I'd like to get all solution angles in some $2\pi$-wide interval. You'd consider any $n\in\mathbb Z$ for which
$$
-1\le\frac{2n+1}{10\sqrt2}\le1\\
-7.57\approx\frac{-10\sqrt2-1}2\le n\le\frac{10\sqrt2-1}2\approx6.57
$$
which matches the list in my original table of solutions.
Your range considerations
But the basic condition of using the above result is that $\beta$ lies between $\left(-\frac π2,\frac π2\right)$.
I'm not sure where you get this condition from. Neither the move from $\cot$ to $\tan$ nor the approach for solving $\tan\theta=\tan\beta$ does warrant such a restriction, as far as I can reason about it.
And so gives $\sin \alpha $ lies between $\left(0,\frac 15\right)$
Since some of the solutions in the above table are outside that range and appear to be valid, that's not the case.
Best Answer
I think this might be what you're looking for. $$\sin(1)\approx\sin(57^\circ)\approx\sin(60^\circ)$$ Infact, this approximation has an error of $3\%$. (Keep in mind that this approximation overestimates.)
Then, you see our final expression becomes $\cos(\sqrt{3}/2)$, i.e. $\cos(0.866)$. Notice that $$\dfrac{\pi}{4}<0.866<\dfrac{\pi}{3}$$ implying ($E$ is our expression $\cos(\sin 1)$). $$\dfrac{\sqrt{2}}{2}>E>\dfrac12$$ Now, you notice that $\dfrac12\approx\dfrac{\pi}{6}$ (error $4\%$). Also, $\pi/4\approx 0.78$ so you're sure that $E<\pi/4$. This was essentially your problem. We don't know if $E$ be less than $\pi/6$, being greater than $1/2$.
Well, in such cases, either you can just look at the options and mark your answer. Or you look at your inequalities and see that $0.866$ lies somewhere in between and not close to one bound. So, your final inequality, $E$ must also be in between and so it isn't harm to replace the bound by something just greater than it. Or you can also see that you overestimated in your first line and so in the final one, the actual quantity should be slightly closer to $\sqrt{2}/2$.
Hope this helps. Ask anything if not clear :)