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Virtual Math Learning Center Texas A&M University Virtual Math Learning Center
tags:
Factoring Quadratic Secant Trigonometry Trigonometry Series

Solving Trigonometric Equations Exercise 3

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all values of \(x\) satisfying \(\sec^2{x}+\sec{x}=2.\)

Solution: \(x=2\pi n,\) \(n \in \mathbb{Z},\quad\) \(x= \dfrac{2\pi}{3}+2\pi n,\) \(n \in \mathbb{Z},\quad\) and  \(x=\dfrac{4\pi}{3}+2\pi n,\) \(n \in \mathbb{Z}\quad \) (\(n \in \mathbb{Z}\) means \(n\) is any integer)

Solution Method: This equation is a "quadratic in secant". If we just subtract the two over to set the LHS equal to zero, we have
\[
\sec^2{x}+\sec{x}-2=0
\]
So it's in the form of \(ax^2+bx+c=0\), but the variable is \(\sec{x}\). Since the equation is in a quadratic form, I can factor it like I would the quadratic equation \(x^2+x-2=0.\) The equation factors to
\begin{align*}
x^2+x-2&=0 \\
(x-1)(x+2) &=0
\end{align*}
So the quadratic in secant factors to
\begin{align*}
\sec^2{x}+\sec{x}-2&=0 \\
(\sec{x}-1)(\sec{x}+2) &= 0
\end{align*}
Then I have two factors to set equal to 0.
The first factor gives us
\begin{align*}
\sec{x}-1 &= 0 \\
\sec{x} &= 1
\end{align*}
Since secant is the reciprocal of cosine, if \(\sec{x}=1\) then
\begin{equation*}
\cos{x} = \frac{1}{1} = 1
\end{equation*}
1 is a quadrantal value, and thinking back to the unit circle I know that \(x=0, 2\pi\) satisfies \(\cos{x} = 1 \). However, the question asks us to find all values of x satisfying the equation, not just the ones in \([0,2\pi]\). Trig functions are periodic; they repeat themselves every \(2\pi\). We can see that in the graphs of sine, cosine, and tangent, and on the unit circle if we keep going around it. So if 0 and \(2\pi\) satisfy the equation, so will \(4\pi, 6\pi, 8\pi...\) and \(-2\pi,-4\pi,...\) See the pattern? They are all integer multiples of \(2\pi\). Once we find this pattern, instead of listing all of these values which are, in fact, infinite, we can write that the values \(x=2\pi n\), for all \(n \in \mathbb{Z}\) satisfy the equation.

Now for the second factor.
\begin{align*}
\sec{x}+2 &= 0 \\
\sec{x} &= -2
\end{align*}
Since secant is the reciprocal of cosine, if \(\sec{x}=-2\) then
\begin{equation*}
\cos{x} = \frac{1}{-2} = -\frac{1}{2}
\end{equation*}
The reference angle that produces a cosine value of \(\frac{1}{2}\) is \(\frac{\pi}{3}\). Then the quadrants in which cosine is negative are (thinking about ASTC) Q2 and Q3. So the angle values satisfying \(\cos{x} = -\frac{1}{2}\) between 0 and \(2\pi\) are \(x=\frac{2\pi}{3}, \frac{4\pi}{3}\). Once again we need to find all values of x satisfying the equation. So we must account for the fact that each full rotation of \(2\pi\) lands us back at an x value satisfying the equation. So \(x=\frac{2\pi}{3}+2\pi n, n\in\mathbb{Z}\) and \(x=\frac{4\pi}{3}+2\pi n, n\in\mathbb{Z}\) also satisfy the equation. The integer values multiplied by the \(2\pi\) represent full rotations.

Note: The equations representing all possible x values can be written in multiple ways. For example, \(x=\frac{2\pi}{3}+2\pi n, n\in\mathbb{Z}\) could also be written as \(x=\frac{1}{3}(2\pi+6\pi n), n\in\mathbb{Z}\). Or since \(\frac{4\pi}{3} = - \frac{2\pi}{3}, x=\frac{4\pi}{3}+2\pi n, n\in\mathbb{Z}\) could be written \(x=-\frac{2\pi}{3}+2\pi n, n\in\mathbb{Z}\)}

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