Answer:
The solutions are described below:
[tex]x_{1} = \frac{\pi}{6} \pm 2\pi\cdot i[/tex] or [tex]x_{1} = \frac{5\pi}{6} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
[tex]x_{2} = \frac{3\pi}{2} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
Step-by-step explanation:
The trigonometric expression must be simplified into a form in terms of one trigonometric function:
[tex]\cos x \tan x - 2\cdot \cos ^{2}x = 1[/tex]
[tex]\sin x - 2\cdot (1 - \sin ^{2}x) = -1[/tex]
[tex]\sin x - 2 + 2\cdot \sin^{2}x = -1[/tex]
[tex]2\cdot \sin^{2}x + \sin x -1=0[/tex]
Roots are determined with the help of the General Equation for the Second-Order Polynomial:
[tex]\sin x = \frac{-1\pm \sqrt{1^{2}-4\cdot (2)\cdot (-1)}}{2\cdot (2)}[/tex]
[tex]\sin x_{1} = \frac{1}{2}[/tex] and [tex]\sin x_{2} = -1[/tex]
The solutions are described below:
[tex]x_{1} = \frac{\pi}{6} \pm 2\pi\cdot i[/tex] or [tex]x_{1} = \frac{5\pi}{6} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
[tex]x_{2} = \frac{3\pi}{2} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
