Can I find a better solution of $\sin(x^2 + x) = \frac 1 2$ for $x$?

Question

It's obvious that $$x^2 + x = (-1)^k \arcsin \frac 1 2 + \pi k$$ where $k \in \Bbb Z$.

If I treat $(-1)^k \arcsin \frac 1 2 + \pi k$ as a constant, I get the answer: $$x_{1, 2} = \frac {-1 \pm \sqrt{1 + 4(-1)^k \arcsin \frac 1 2 + \pi k}} {2}$$ The ugliness and length of the answer makes me think I did something wrong.

Background: Knowing the structure of my textbook, I think authors didn't expect me to use addition and subtraction theorems. Also note that this is a preparation for a college entrance exam.

Answer 1

Why not treat $(\frac\pi6+2k\pi)$and $(\frac{5\pi}6+2k\pi)$ as a constant and use the quadratic formula?

Answer 2

OK, I found out that this is actually a wrong answer since I say that $k \in \Bbb Z$.

It made me no good that I shortened the answer to a single line. Let's take a look at the expanded version: $$ x = \frac {-1 \pm \sqrt{1 + \frac \pi 6 + 2 \pi k}} {2} \tag{1}$$ $$ x = \frac {-1 \pm \sqrt{1 + \frac {5 \pi} 6 + 2 \pi k}} {2} \tag{2}$$ Now I can see that there are certain restrictions: $$ 1 + \frac \pi 6 + 2 \pi k \ge 0 $$ $$ 1 + \frac {5 \pi} 6 + 2 \pi k \ge 0 $$ For $k \ge 0$, the inequality holds. One can check that this is not true for $k \le -1$.

So the answer is the expressions $(1), (2)$ where $k = 0, 1, 2, 3, …$

So yes, I can find a better solution (because the right solution is better than the wrong one). I suppose this was the catch of the exercise.