Solving 4 Sin^2 Θ - 1 = 0 On The Interval 0 ≤ Θ < 2π
In the realm of mathematics, particularly in trigonometry, solving trigonometric equations is a fundamental skill. These equations, which involve trigonometric functions such as sine, cosine, and tangent, often appear in various scientific and engineering applications. In this article, we will delve into a detailed explanation of how to solve the trigonometric equation 4 sin^2 θ - 1 = 0 within the interval 0 ≤ θ < 2π. This interval represents one full revolution around the unit circle, and finding solutions within this range is crucial for many practical problems. Understanding the methods and concepts presented here will not only help you solve this specific equation but also equip you with the tools to tackle a wide range of trigonometric problems.
Before diving into the solution, let's first understand what trigonometric equations are and why they are important. Trigonometric equations are equations that involve trigonometric functions of an unknown angle, which we often denote as θ (theta). These equations are different from trigonometric identities, which are true for all values of the angle. Instead, trigonometric equations are true only for specific values of the angle. Solving these equations means finding all the angles that satisfy the given equation within a specified interval, which in our case is 0 ≤ θ < 2π.
Trigonometric functions, such as sine (sin), cosine (cos), and tangent (tan), relate the angles of a right triangle to the ratios of its sides. These functions are periodic, meaning their values repeat at regular intervals. The sine and cosine functions have a period of 2π, while the tangent function has a period of π. This periodicity is a key aspect when solving trigonometric equations because it means there can be multiple solutions within a given interval. Recognizing and utilizing this periodic behavior is essential for finding all possible solutions.
In many real-world applications, trigonometric equations are used to model periodic phenomena such as oscillations, waves, and rotations. For instance, in physics, these equations can describe the motion of a pendulum or the propagation of light waves. In engineering, they are used in the design of electrical circuits and mechanical systems. Therefore, mastering the techniques for solving trigonometric equations is not just an academic exercise but a valuable skill with practical applications.
Now, let's tackle the equation 4 sin^2 θ - 1 = 0. Our goal is to find all values of θ in the interval 0 ≤ θ < 2π that satisfy this equation. We will approach this problem systematically, breaking it down into manageable steps.
Step 1: Isolate the Trigonometric Function
The first step in solving any trigonometric equation is to isolate the trigonometric function. In our case, we want to isolate sin^2 θ. To do this, we'll perform a series of algebraic manipulations. Starting with the equation 4 sin^2 θ - 1 = 0, we add 1 to both sides:
4 sin^2 θ = 1
Next, we divide both sides by 4:
sin^2 θ = 1/4
Now we have successfully isolated sin^2 θ on one side of the equation. This simplification is crucial because it allows us to proceed with the next steps more easily.
Step 2: Take the Square Root
Once we have isolated sin^2 θ, the next step is to take the square root of both sides of the equation. Remember that when taking the square root, we need to consider both the positive and negative roots. This is because both the positive and negative values, when squared, will give the same result. So, taking the square root of sin^2 θ = 1/4 gives us:
sin θ = ±√(1/4)
Simplifying the square root, we get:
sin θ = ±1/2
This result tells us that we are looking for angles θ where the sine function is either 1/2 or -1/2. This is a critical step because it breaks the problem into two simpler equations: sin θ = 1/2 and sin θ = -1/2.
Step 3: Find the Reference Angles
To find the angles θ, we need to determine the reference angles. A reference angle is the acute angle formed between the terminal side of the angle and the x-axis. It helps us find all the angles in different quadrants that have the same sine value. For sin θ = 1/2, the reference angle is the angle whose sine is 1/2. This angle is:
θ_ref = π/6
This is a standard angle that you should be familiar with from the unit circle or trigonometric tables. Now, we need to find the angles in the interval 0 ≤ θ < 2π that have this reference angle and satisfy the conditions sin θ = 1/2 and sin θ = -1/2.
Step 4: Determine the Quadrants
The next step is to determine in which quadrants the solutions lie. The sine function is positive in the first and second quadrants and negative in the third and fourth quadrants. This is based on the unit circle, where the y-coordinate corresponds to the sine value.
For sin θ = 1/2, we need to find angles in the first and second quadrants that have a reference angle of π/6.
For sin θ = -1/2, we need to find angles in the third and fourth quadrants that have a reference angle of π/6.
Step 5: Find the Solutions
Now we can find the solutions in each quadrant.
For sin θ = 1/2:
- Quadrant I: In the first quadrant, the angle is simply the reference angle:
θ_1 = π/6
- Quadrant II: In the second quadrant, the angle is given by:
θ_2 = π - θ_ref = π - π/6 = 5π/6
For sin θ = -1/2:
- Quadrant III: In the third quadrant, the angle is given by:
θ_3 = π + θ_ref = π + π/6 = 7π/6
- Quadrant IV: In the fourth quadrant, the angle is given by:
θ_4 = 2π - θ_ref = 2π - π/6 = 11π/6
So, the solutions to the equation 4 sin^2 θ - 1 = 0 in the interval 0 ≤ θ < 2π are π/6, 5π/6, 7π/6, and 11π/6.
Finally, we can express the solution set. The solution set is the set of all angles θ that satisfy the equation within the given interval. In this case, the solution set is:
{π/6, 5π/6, 7π/6, 11π/6}
This set contains all the angles between 0 and 2π where 4 sin^2 θ - 1 = 0. These are the angles where the sine function is either 1/2 or -1/2, as we determined earlier.
While we have solved the equation using a step-by-step algebraic approach, there are alternative methods and insights that can deepen our understanding of trigonometric equations.
Using the Unit Circle
The unit circle is a powerful tool for visualizing and solving trigonometric equations. It is a circle with a radius of 1 centered at the origin in the Cartesian plane. The coordinates of any point on the unit circle are given by (cos θ, sin θ), where θ is the angle measured counterclockwise from the positive x-axis.
To solve 4 sin^2 θ - 1 = 0 using the unit circle, we look for points on the circle where the y-coordinate (sin θ) is either 1/2 or -1/2. By visualizing the circle, we can quickly identify the angles π/6, 5π/6, 7π/6, and 11π/6 as the points where the y-coordinate meets these conditions. This method provides a visual confirmation of our algebraic solution and can be particularly useful for more complex equations.
Factoring the Equation
Another approach to solving 4 sin^2 θ - 1 = 0 is by factoring. We can rewrite the equation as a difference of squares:
(2 sin θ)^2 - 1^2 = 0
Using the difference of squares factorization, a^2 - b^2 = (a - b)(a + b), we get:
(2 sin θ - 1)(2 sin θ + 1) = 0
This gives us two separate equations:
2 sin θ - 1 = 0 or 2 sin θ + 1 = 0
Solving these equations individually:
sin θ = 1/2 or sin θ = -1/2
This method directly leads us to the same trigonometric equations we derived earlier, and we can proceed to find the solutions as before. Factoring can be a valuable technique for solving trigonometric equations, especially when they can be expressed in a factorable form.
General Solutions
While we have found the solutions in the interval 0 ≤ θ < 2π, it's important to note that the trigonometric functions are periodic. This means there are infinitely many solutions to the equation if we consider all possible angles. The general solutions can be expressed by adding integer multiples of the period (2π for sine) to the solutions we found in the interval 0 ≤ θ < 2π.
For sin θ = 1/2, the general solutions are:
θ = π/6 + 2πk
θ = 5π/6 + 2πk
where k is any integer.
For sin θ = -1/2, the general solutions are:
θ = 7π/6 + 2πk
θ = 11π/6 + 2πk
These general solutions represent all possible angles that satisfy the equation, not just those in the interval 0 ≤ θ < 2π.
Solving trigonometric equations can be challenging, and there are several common mistakes that students often make. Being aware of these pitfalls can help you avoid them and improve your accuracy.
Forgetting the ± Sign
One of the most common mistakes is forgetting to consider both the positive and negative roots when taking the square root. For example, when solving sin^2 θ = 1/4, it's crucial to remember that sin θ can be either 1/2 or -1/2. Neglecting the negative root will lead to missing solutions.
To avoid this, always remember to include the ± sign when taking the square root in an equation.
Incorrectly Identifying Quadrants
Another common error is misidentifying the quadrants where the solutions lie. Remember that sine is positive in the first and second quadrants and negative in the third and fourth quadrants. Cosine is positive in the first and fourth quadrants and negative in the second and third quadrants. Tangent is positive in the first and third quadrants and negative in the second and fourth quadrants.
To avoid quadrant errors, use the mnemonic **
