91Ó°ÊÓ

Quadratic equations are a very common type of equation in algebra and have the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. A quadratic equation can have up to two real solutions. These solutions are usually found using various methods such as factoring, the quadratic formula, or completing the square.

In our example, we began by recognizing the trigonometric equation \(\sec^2 x - \sec x - 2 = 0\) could be rewritten as a quadratic equation by substituting \(u = \sec x\). This gave us the quadratic \(u^2 - u - 2 = 0\).

To solve a quadratic, you can look for two numbers that multiply to the constant term \(c\), and add up to the linear coefficient \(b\). Here, the equation factors neatly to \((u-2)(u+1) = 0\), giving the solutions \(u = 2\) and \(u = -1\). These solutions for \(u\), when mapped back to \(x\), help find the solutions to the original trigonometric equation.

Trigonometry often involves using identities to simplify or solve equations. An identity is an equation that is true for all values within its domain. In our exercise, we used the trigonometric identity \(\sec x = \frac{1}{\cos x}\).

This identity is crucial because it connects the secant function with the cosine function, allowing us to convert between the two. Once the quadratic solutions were found as \(u = 2\) and \(u = -1\), we needed to transition back to the original variable \(x\) using the identity.

Thus, for \(\sec x = 2\), we solve \(\cos x = \frac{1}{2}\). Similarly, \(\sec x = -1\) translates to \(\cos x = -1\). Knowing the trigonometric circle and common values for cosine helps identify the corresponding angles that satisfy these equations within a specified interval.

Interval notation provides a way to describe ranges of solutions. It is often used in mathematics to specify the domain or range of a function, or to communicate where particular solutions are valid.

In our exercise, we are tasked with finding solutions within the interval \([0, 2\pi)\). This notation indicates all values from 0 to just before \(2\pi\) (exclusive of \(2\pi\)). The use of a square bracket \([\) shows that 0 is included, while a parenthesis \()\) denotes that \(2\pi\) is not included.

By solving \(\sec x = 2\) and knowing \(\cos x = \frac{1}{2}\) in the interval, we find \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\). Similarly, solving \(\sec x = -1\) translates to \(\cos x = -1\), giving us the solution \(x = \pi\). These results are all within \([0, 2\pi)\), making them valid solutions for the original trigonometric equation.