Answer :

To solve the equation [tex]\(\sin^2 \theta - \sin \theta - 6 = 0\)[/tex], we'll follow these detailed steps:

1. Substitute: Let [tex]\( x = \sin \theta \)[/tex]. This substitution changes our equation in terms of [tex]\(\sin \theta\)[/tex] into a quadratic equation in [tex]\( x \)[/tex]:
[tex]\[ x^2 - x - 6 = 0 \][/tex]

2. Solve the quadratic equation:
The quadratic equation [tex]\( x^2 - x - 6 = 0 \)[/tex] can be solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex]. Plugging these values into the quadratic formula, we get:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{1 + 5}{2} = 3 \quad \text{and} \quad x = \frac{1 - 5}{2} = -2 \][/tex]

3. Back-substitute: Recall [tex]\( x = \sin \theta \)[/tex]:
[tex]\[ \sin \theta = 3 \quad \text{and} \quad \sin \theta = -2 \][/tex]

4. Evaluate the solutions:
- [tex]\(\sin \theta = 3\)[/tex] is not a valid solution because the sine function only takes values in the range [tex]\([-1, 1]\)[/tex].
- [tex]\(\sin \theta = -2\)[/tex] is similarly not a valid solution for the same reason.

Therefore, the given equation [tex]\(\sin^2 \theta - \sin \theta - 6 = 0\)[/tex] has no real solutions that correspond to valid sine values.

Given the context, understanding the apparent discrepancy can reveal potential deeper insights where complex solutions or alternative domains may be involved, corresponding to solutions involving arcsine functions in specific contexts (not typical in real number scenarios). The expressions such as [tex]\(\pi + \arcsin(2)\)[/tex], [tex]\(\pi - \arcsin(3)\)[/tex], [tex]\(-\arcsin(2)\)[/tex], [tex]\(\arcsin(3)\)[/tex] are evaluated in extended complex domains respecting the general approach in the initial aspect but here in real numbers, such solutions do not exist.