Problem 15 - Precalculus with Limits

Question

In Exercises 11-24, solve the equation. \( 3 \sec^2 x - 4 = 0 \)

Step-by-Step Solution

Answer

The solutions of the equation are \(x = \pm\frac{\pi}{6}+2\pi k\) and \(x = \pm\frac{5\pi}{6}+2\pi k\), where k is an integer.

Step 1: Isolate the secant term

We start by adding 4 to both sides of the equation, giving us \(3 \sec^2x = 4\).

Step 2: Simplify the isolated secant term

Divide both sides of the equation by 3, which simplifies to \(\sec^2x = \frac{4}{3}\).

Step 3: Take the square root

Taking the square root of both sides, realizing that x can be positive or negative, gives us \(\sec x = \pm\sqrt{\frac{4}{3}}\).

Step 4: Convert secant to cosine

We know that sec(x) = 1/cos(x). Therefore, we can convert the equation to a form that uses cosine, giving us \( \cos x = \pm\frac{1}{\sqrt{\frac{4}{3}}}\).

Step 5: Simplify and solve for x

Simplify the fraction to \( \cos x = \pm\frac{\sqrt{3}}{2}\). The values of x that satisfy the equation are \( x = \pm\frac{\pi}{6}+2\pi k\) and \(x = \pm\frac{5\pi}{6}+2\pi k\), where k is an integer.

Key Concepts

secant function

The secant function, denoted as \( ext{sec}(x) \), is a fundamental trigonometric function. It is defined as the reciprocal of the cosine function. This means that \( ext{sec}(x) = \frac{1}{\cos(x)} \).
The secant function is important in trigonometry because it helps us understand the behavior of angles in right triangles and circles.
For any angle \( x \), the secant function can reach very large values as cosine approaches zero. This is because dividing by a small number (close to zero) results in a large number.

• Secant is undefined where \( ext{cos}(x) = 0 \).
• It's periodic, like other trigonometric functions, with a period of \( 2\pi \).

Understanding the secant function allows us to solve trigonometric equations that involve reciprocal relationships.

cosine function

The cosine function, represented as \( ext{cos}(x) \), is another core trigonometric function. It relates to the coordinates of a point on the unit circle.
Let's talk about its essentials:

• The cosine of an angle in a right triangle represents the ratio of the adjacent side to the hypotenuse.
• It has a range of values from -1 to 1 and is periodic with a period of \( 2\pi \).
• Cosine is an even function, meaning \( ext{cos}(-x) = ext{cos}(x) \).

In equations, cosine is very commonly used, and its relationship with secant (being its reciprocal) is utilized to solve problems. Understanding how to manipulate the cosine in equations, such as changing its form, is crucial for solving trigonometric equations.

solving trigonometric equations

Solving trigonometric equations involves finding all angle values that satisfy a given trigonometric expression.
Here is a simple process to follow:

• First, isolate the trigonometric function involved.
• Use algebraic methods to manipulate the equation, simplifying where necessary.
• Utilize trigonometric identities to transform the expression if needed.
• Find the general solutions, considering the periodic nature of trig functions.

In practice, like in the exercise where \( 3 \sec^2(x) - 4 = 0 \) was solved, each step brought us closer to identifying solutions. First, the equation was simplified to \( \sec(x) = \pm\sqrt{\frac{4}{3}} \), and then converted to cosine values for more straightforward solutions.

trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are always true for their input angles. These identities are indispensable tools in solving trigonometric problems.
Below are a few commonly used identities and their applications:

• Pythagorean identity: \( \sin^2(x) + \cos^2(x) = 1 \)
• Reciprocal identities: \( ext{sec}(x) = \frac{1}{\cos(x)} \) and \( ext{csc}(x) = \frac{1}{\sin(x)} \)
• Angle sum identities: \( \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) \)

In our example, we used the reciprocal identity where \( ext{sec}(x) = \frac{1}{\cos(x)} \) to convert the secant equation into one with cosine. This strategy enabled us to solve the equation efficiently.