Trigonometric equations are equations involving trigonometric functions such as sine, cosine, and tangent. These equations seek to find specific values, known as solutions, that satisfy the equation.
Trigonometric functions are inherently periodic, meaning they repeat their values at regular intervals. As a result, trigonometric equations may have multiple solutions due to this periodic nature. In fact, some equations may have infinitely many solutions.
To address all possible solutions, we use a technique known as a general solution. This method involves initially identifying solutions within a single period of the trigonometric function. We then extend these solutions by adding integer multiples of the period of the trigonometric function.
In this section, we will explore various techniques for effectively solving trigonometric equations, including methods for finding general solutions.
Subsection3.5.1Solving Equations with a Single Trigonometric Function
While this is a valid solution, it’s important to recognize that there are additional solutions to consider.
Since the sine function is positive in both Quadrant I and Quadrant II, we can find another solution in Quadrant II by using the reference angle \(\frac{\pi}{6}\) and the methods described in Subsection 1.5.2. Therefore, the equivalent angle in Quadrant II is \(\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\text{.}\)
However, these two angles, \(\theta=\frac{\pi}{6}\) or \(\theta=\frac{5\pi}{6}\text{,}\) are not the only solutions. Recall that the sine function has a period of \(2\pi\text{,}\) meaning that adding or subtracting any integer multiples of \(2\pi\) to these angles will also give you solutions. For example, \(\theta=\frac{\pi}{6}+4\pi\) and \(\theta=\frac{5\pi}{6}-10\pi\) are both solutions.
Thus, the general solution to \(\sin\theta=\frac{1}{2}\) can be expressed as:
From Table 1.5.18, when \(\cos\theta=\frac{\sqrt{2}}{2}\text{,}\) we have \(\theta=\frac{\pi}{4}\) or \(\theta=\frac{7\pi}{4}\text{;}\) and when \(\cos\theta=-\frac{\sqrt{2}}{2}\text{,}\) we have \(\theta=\frac{3\pi}{4}\) or \(\theta=\frac{5\pi}{4}\text{.}\)
Subsection3.5.3Solving Trigonometric Equations by Factoring
Example3.5.4.
Solve \(\cos^2x - 4\cos x + 3 = 0\text{.}\)
Solution.
To solve this equation, let’s make a substitution to simplify it. We’ll let \(y = \cos x\text{,}\) so the equation becomes \(y^2 - 4y + 3 = 0\text{.}\)
Factoring the quadratic equation, we obtain \((y - 3)(y - 1) = 0\text{.}\) Setting each factor equal to zero, we find two potential solutions: \(y = 3\) or \(y = 1\text{.}\) We are not done because we need to solve for \(x\) and not \(y\text{.}\)
Substituting back \(\cos x\) for \(y\text{,}\) we find that \(\cos x = 3\) is not a valid solution, as the range of cosine is limited to [-1, 1]. However, \(\cos x = 1\) yields a solution of \(x = 0\) for one period.
Therefore, the general solution to the equation is:
\begin{equation*}
x = 0+2k\pi = 2k\pi,
\end{equation*}
Rounding to two decimals, we get \(\theta=0.59\) radians, which is in Quadrant I since \(0\lt0.59\lt\frac{\pi}{2}\text{.}\) Another Quadrant where \(\tan\theta=\frac{2}{3}\) is in Quadrant III. Using the methods in Subsection 1.5.2 we get the other angle: \(\theta=0.59+\pi\text{.}\)
Exercises3.5.5Exercises
Exercise Group.
Solve each equation on the interval \(0\leq\theta\lt2\pi\)