Answer:
a) [tex]x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex], [tex]x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex], b) [tex]x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex], [tex]x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
Step-by-step explanation:
a) The equation must be rearranged into a form with one fundamental trigonometric function first:
[tex]\sqrt{3}\cdot \csc x - 2 = 0[/tex]
[tex]\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0[/tex]
[tex]\sqrt{3} - 2\cdot \sin x = 0[/tex]
[tex]\sin x = \frac{\sqrt{3}}{2}[/tex]
[tex]x = \sin^{-1} \frac{\sqrt{3}}{2}[/tex]
Value of x is contained in the following sets of solutions:
[tex]x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
[tex]x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
b) The equation must be simplified first:
[tex]\cos x + 1 = - \cos x[/tex]
[tex]2\cdot \cos x = -1[/tex]
[tex]\cos x = -\frac{1}{2}[/tex]
[tex]x = \cos^{-1} \left(-\frac{1}{2} \right)[/tex]
Value of x is contained in the following sets of solutions:
[tex]x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
[tex]x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
