Question

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Answer 1

Answer:

$x = 0$ or $x = \pi$.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$.

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$. Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$.

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$.

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$. Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$.

$\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2$.

$\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2$.

Solve this equation for $u$:

$\displaystyle \frac{u^{2} + 1}{u^{2}} = 2$.

$u^{2} + 1 = 2 u^{2}$.

$u^{2} = 1$.

Given that $0 \le u \le 1$, $u = 1$ is the only possible solution.

$\cos^{2}{x} = 1$,

$x = k \pi$, where $k\in \mathbb{Z}$ (i.e., $k$ is an integer.)

Given that $0 \le x < 2\pi$,

$0 \le k$.

$k = 0$ or $k = 1$. Accordingly,

$x = 0$ or $x = \pi$.

Answer 2

Answer:

Step-by-step explanation: