Answer:
[tex]x \neq 0 \pm \pi \cdot i, \forall i \in \mathbb{N}_{O}[/tex] and [tex]x \neq \frac{3\pi}{2}\pm 2\pi \cdot j, \forall j \in \mathbb{N}_{O}[/tex].
Step-by-step explanation:
The equation can be simplified with the help of trigonometric identities:
[tex]\sin x \cdot (\sin x + 1) = 0[/tex]
Which means that equation is equal to zero if any component of the product is zero. The solutions of the expression are:
a) [tex]\sin x[/tex]
[tex]x = 0 \pm \pi\cdot i,\forall i \in \mathbb{N}_{O}[/tex]
b) [tex]\sin x + 1[/tex]
[tex]\sin x + 1 = 0[/tex]
[tex]\sin x = -1[/tex]
[tex]x = \frac{3\pi}{2} \pm 2\pi\cdot i,\forall i \in \mathbb{N}_{O}[/tex]
Any value different of both subsets do not satisfy the equation described above.
