Solving Sec²(x) ≤ Tan²(x) + Sec(x) Over 0 To 2π

This article delves into the process of solving the trigonometric inequality $\sec^2(x) \leq \tan^2(x) + \sec(x)$ within the interval $0 \leq x < 2\pi$. We'll explore the trigonometric identities involved, the steps to isolate the variable, and the determination of the solution set. This detailed explanation aims to provide a comprehensive understanding of the solution, making it accessible to students and enthusiasts alike.

Understanding the Problem

Before diving into the solution, let's first understand the trigonometric functions involved and the given inequality. The key trigonometric functions in this problem are secant (sec) and tangent (tan), which are defined as follows:

• $\sec(x) = \frac{1}{\cos(x)}$
• $\tan(x) = \frac{\sin(x)}{\cos(x)}$

The given inequality is $\sec^2(x) \leq \tan^2(x) + \sec(x)$. Our goal is to find all values of $x$ in the interval $[0, 2\pi)$ that satisfy this inequality. To achieve this, we will use trigonometric identities to simplify the inequality and isolate the variable. The interval $[0, 2\pi)$ represents one full rotation around the unit circle, which is important for finding all possible solutions.

Utilizing Trigonometric Identities

To solve the inequality, we can leverage the fundamental Pythagorean trigonometric identity: $\sin^2(x) + \cos^2(x) = 1$. Dividing both sides of this identity by $\cos^2(x)$, we obtain another crucial identity:

$\frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}$

This simplifies to:

$\tan^2(x) + 1 = \sec^2(x)$

This identity, $\tan^2(x) + 1 = \sec^2(x)$, is a cornerstone for simplifying our inequality. Substituting this identity into the original inequality, $\sec^2(x) \leq \tan^2(x) + \sec(x)$, allows us to eliminate the squared terms and work with a simpler expression. This substitution is a common technique in solving trigonometric inequalities, as it reduces the complexity and makes the inequality easier to manipulate. By replacing $\sec^2(x)$ with $\tan^2(x) + 1$, we pave the way for isolating the secant function and ultimately finding the solution set.

Simplifying the Inequality

Now, let's substitute the identity $\sec^2(x) = \tan^2(x) + 1$ into the inequality:

$\tan^2(x) + 1 \leq \tan^2(x) + \sec(x)$

Subtracting $\tan^2(x)$ from both sides, we get:

$1 \leq \sec(x)$

This simplified inequality is much easier to work with. It tells us that we need to find the values of $x$ for which the secant of $x$ is greater than or equal to 1. Recalling that $\sec(x) = \frac{1}{\cos(x)}$, we can rewrite the inequality as:

$1 \leq \frac{1}{\cos(x)}$

This form allows us to directly analyze the behavior of the cosine function, which is more intuitive for many. Understanding the relationship between secant and cosine is crucial here. Secant is the reciprocal of cosine, so when secant is greater than or equal to 1, cosine must be less than or equal to 1. This reciprocal relationship is a fundamental concept in trigonometry and is key to solving this problem efficiently. Now, we need to determine the intervals where $\cos(x)$ satisfies this condition within the given range of $x$.

Analyzing the Cosine Function

To solve the inequality $1 \leq \frac{1}{\cos(x)}$, we can take the reciprocal of both sides. However, we need to be cautious about the sign. Since we are dealing with an inequality, we must consider when $\cos(x)$ is positive or negative. If $\cos(x)$ is positive, taking the reciprocal will flip the inequality sign. If $\cos(x)$ is negative, the inequality sign will remain the same. However, since $1 \leq \frac{1}{\cos(x)}$, it implies that $\cos(x)$ must be positive (or undefined when $\cos(x) = 0$). Thus, we have:

$\cos(x) \leq 1$

This inequality is true for all $x$, except where $\cos(x)$ is undefined, which occurs when $\cos(x) = 0$. The cosine function equals zero at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ within the interval $[0, 2\pi)$. Therefore, these values must be excluded from our solution set. This is because the original inequality involves $\sec(x)$, which is undefined when $\cos(x) = 0$. Understanding the behavior of the cosine function and identifying its zeros is essential for correctly determining the solution intervals. By excluding the points where cosine is zero, we ensure that our solution remains valid within the context of the original inequality.

Addressing the Undefined Points

The original inequality $\sec^2(x) \leq \tan^2(x) + \sec(x)$ involves $\sec(x)$ and $\tan(x)$, both of which are undefined when $\cos(x) = 0$. This occurs at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ within the interval $[0, 2\pi)$. Therefore, these values must be excluded from the solution set. These undefined points represent vertical asymptotes for the secant and tangent functions, which means the functions approach infinity at these points. Including these points in the solution would lead to mathematical inconsistencies. By recognizing and excluding these points, we ensure that the solution is mathematically sound and accurately reflects the behavior of the trigonometric functions involved. This attention to detail is crucial for solving trigonometric inequalities correctly.

Determining the Solution Set

Since $\cos(x) \leq 1$ is true for all $x$, but $\sec(x)$ is undefined at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$, the solution set for the inequality $\sec^2(x) \leq \tan^2(x) + \sec(x)$ over the interval $[0, 2\pi)$ is:

$0 \leq x < \frac{\pi}{2}$ and $\frac{3\pi}{2} < x < 2\pi$

This solution set represents the intervals where the inequality holds true, excluding the points where the functions are undefined. The solution can be visualized on the unit circle, where these intervals correspond to the angles for which the secant function is greater than or equal to 1. Understanding the geometric interpretation of the solution can provide a deeper insight into the behavior of the trigonometric functions and the inequality itself. By combining algebraic manipulation with graphical understanding, we can confidently arrive at the correct solution set.

Conclusion

In conclusion, solving the trigonometric inequality $\sec^2(x) \leq \tan^2(x) + \sec(x)$ over the interval $[0, 2\pi)$ involves utilizing trigonometric identities, simplifying the inequality, analyzing the behavior of trigonometric functions, and carefully addressing undefined points. The solution set is $0 \leq x < \frac{\pi}{2}$ and $\frac{3\pi}{2} < x < 2\pi$. This process demonstrates the importance of a solid understanding of trigonometric functions and identities in solving mathematical problems. By systematically applying these concepts, we can successfully navigate complex inequalities and arrive at accurate solutions. This detailed approach not only solves the specific problem but also enhances our overall problem-solving skills in mathematics.