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
The OP finds one constant root that applies for all $a$. But there is a second root for most specific values of $a$, which is a function of $a$. The full answer is $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.
Properly, the given equation should be combined with the identity $\sin^2\alpha+\cos^2\alpha=1$. There are two ways to do this:
Method 1
Isolate one of the trigonometric function, square the resulting equations and substitute to get a quadratic equation for the remaining function. Choosing to isolate the cosine we then have
$(2a-1)\cos\alpha=(2a+1)-(a+2)\sin\alpha$
$(2a-1)^2\cos^2\alpha=(2a+1)^2-2(2a+1)(a+2)\sin\alpha+(a+2)^2\sin^2\alpha$
$(4a^2-4a+1)-(4a^2-4a+1)\sin^2\alpha=(4a^2+4a+1)-(4a^2+10a+4)\sin\alpha+(a^2+4a+4)\sin^2\alpha$
$(5a^2+5)\sin^2\alpha-(4a^2+10a+4)\sin\alpha+8a=0$
The quadratic equation looks like a mouthful, but its discriminant is a squared quantity, to wit $(4a^2-10a+4)^2$, thus we get the two roots
$\sin\alpha=\dfrac{(4a^2+10a+4)\pm(4a^2-10a+4)}{2(5a^2+5)}\in\{4/5,2a/(a^2+1)\}$
For each root of $\sin\alpha$ the previous equation with $\cos\alpha$ isolated is used to assure the proper sign of that function:
$(2a-1)\cos\alpha=(2a+1)-(a+2)(4/5); \cos\alpha=3/5$
$(2a-1)\cos\alpha=(2a+1)-(a+2)(2a/(a^2+1)); \cos\alpha=(a^2-1)/(a^2+1)$
Correspondingly $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.
Thus the answer given by the OP is correct for one root that applies for all $a$, but in most cases there will be a second root for any specific value of $a$ (the only exceptions being $a=2$ where there is one doubly degenerate root instead, and $a=\pm 1$ where the second root fails to give a defined value for $\tan\alpha$; also $a=-1/2$ gives the second root with $\tan\alpha=4/3$ but different values for the sine and cosine). The existence of two roots ultimately comes from the fact that except for $\pm 1$, any value of the sine or cosine corresponds to two different arguments within any fundamental period.
Method 2
In this method, we use a trick by combining the original equation with one involving the orthogonal linear combination. To get the orthogonal combination, switch the coefficients $a+2$ and $2a-1$ and then reverse the sign before the second term.
$(a+2)\sin\alpha+(2a-1)\cos\alpha=(2a+1)$
$(2a-1)\sin\alpha-(a+2)\cos\alpha=x$
Square both sides and add them together getting:
$(5a^2+5)(\sin^2\alpha+\cos^2\alpha)=5a^2+5=(2a+1)^2+x^2$
Thus $x=\pm\sqrt{5a^2+5-(2a+1)^2}=\pm(a-2)$. We then have a linear system to solve for $\sin\alpha$ and $\cos\alpha$ for each root of $x$. For example, $x=+(a-2)$ gives
$(a+2)\sin\alpha+(2a-1)\cos\alpha=(2a+1)$
$(2a-1)\sin\alpha-(a+2)\cos\alpha=a-2$
whose solution is $\sin\alpha=4/5,\cos\alpha=3/5$. Putting $-(a-2)$ for $x$ similarly gives a linear system with solution $\sin\alpha=2a/(a^2+1),\cos\alpha=(a^2-1)/(a^2+1)$.
The remainder of the solution, and the comments that follow, are identical to Method 1.