Answer:
[tex]x = 0[/tex] or [tex]x = \pi[/tex].
Step-by-step explanation:
How are tangents and secants related to sines and cosines?
[tex]\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}[/tex].
[tex]\displaystyle \sec{x} = \frac{1}{\cos{x}}[/tex].
Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, [tex]\sin^{2}{x} = 1 - \cos^{2}{x}[/tex]. Therefore, for the square of tangents,
[tex]\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}[/tex].
This equation will thus become:
[tex]\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2[/tex].
To simplify the calculations, replace all [tex]\cos^{2}{x}[/tex] with another variable. For example, let [tex]u = \cos^{2}{x}[/tex]. Keep in mind that [tex]0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1[/tex].
[tex]\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2[/tex].
[tex]\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2[/tex].
Solve this equation for [tex]u[/tex]:
[tex]\displaystyle \frac{u^{2} + 1}{u^{2}} = 2[/tex].
[tex]u^{2} + 1 = 2 u^{2}[/tex].
[tex]u^{2} = 1[/tex].
Given that [tex]0 \le u \le 1[/tex], [tex]u = 1[/tex] is the only possible solution.
[tex]\cos^{2}{x} = 1[/tex],
[tex]x = k \pi[/tex], where [tex]k\in \mathbb{Z}[/tex] (i.e., [tex]k[/tex] is an integer.)
Given that [tex]0 \le x < 2\pi[/tex],
[tex]0 \le k <2[/tex].
[tex]k = 0[/tex] or [tex]k = 1[/tex]. Accordingly,
[tex]x = 0[/tex] or [tex]x = \pi[/tex].
