Question

Math:

Answer 1

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$, $\forall i \in \mathbb{N}_{O}$, $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$, $\forall i \in \mathbb{N}_{O}$, b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$, $\forall i \in \mathbb{N}_{O}$, $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$, $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$, $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$, $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

$x = \cos^{-1} \left(-\frac{1}{2} \right)$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$, $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$, $\forall i \in \mathbb{N}_{O}$