Solving 2sin²x = Sin X A Step-by-Step Guide

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In the realm of trigonometry, solving equations is a fundamental skill. These equations involve trigonometric functions like sine, cosine, tangent, and their reciprocals. The goal is to find the values of the variable (often represented as 'x' or 'θ') that satisfy the equation within a specified interval. Trigonometric equations arise in various applications, from physics and engineering to computer graphics and navigation. Mastering the techniques for solving them is crucial for anyone working with these fields.

Understanding Trigonometric Equations

Trigonometric equations are mathematical statements that involve trigonometric functions, such as sine, cosine, tangent, cotangent, secant, and cosecant. These functions relate angles to the ratios of sides in a right triangle. Solving trigonometric equations means finding the values of the variable (usually an angle) that make the equation true. Unlike algebraic equations that typically have a finite number of solutions, trigonometric equations often have infinitely many solutions due to the periodic nature of trigonometric functions.

For example, consider the equation sin(x) = 0. This equation has solutions at x = 0, x = π, x = 2π, and so on, because the sine function equals zero at these angles. To handle this infinite nature, we often specify an interval within which we seek solutions. A common interval is [0, 2π), which represents one full cycle around the unit circle. Understanding the periodicity and symmetry of trigonometric functions is essential for finding all solutions within a given interval.

The Importance of the Interval [0, 2π)

The interval [0, 2π) is particularly significant in trigonometry because it represents one complete cycle of the trigonometric functions. In radians, 2π corresponds to 360 degrees, which is a full rotation around the unit circle. When we restrict our solutions to this interval, we're essentially finding all unique solutions within one cycle. This is important because the trigonometric functions repeat their values every 2π radians (or 360 degrees). For instance, sin(x) = sin(x + 2π) = sin(x + 4π), and so on.

By focusing on the interval [0, 2π), we can identify the primary solutions and then use the periodicity of the trigonometric functions to find additional solutions if needed. This interval also aligns with the unit circle, where angles are measured counterclockwise from the positive x-axis. The coordinates of points on the unit circle directly correspond to the cosine and sine of the angle, making it a valuable tool for visualizing and solving trigonometric equations. Therefore, solutions within [0, 2π) provide a comprehensive understanding of the equation's behavior within a fundamental cycle.

Solving the Equation 2sin²x = sin x

To illustrate the process of solving trigonometric equations, let's consider the equation 2sin²x = sin x. This equation involves the sine function and requires algebraic manipulation to isolate the variable x. Our goal is to find all solutions for x within the interval [0, 2π). This means we are looking for angles between 0 and 360 degrees (exclusive of 360 degrees) that satisfy the equation. The steps we'll follow include rearranging the equation, factoring, and using the unit circle or trigonometric identities to find the solutions.

Step-by-Step Solution

  1. Rearrange the Equation: The first step in solving 2sin²x = sin x is to set the equation equal to zero. This is a common strategy in solving quadratic-like equations. By moving all terms to one side, we can factor the equation and find the values of sin x that make the equation true. Subtracting sin x from both sides gives us:

    2sin²x - sin x = 0

  2. Factor the Equation: Now that we have the equation in the form 2sin²x - sin x = 0, we can factor out the common factor, which is sin x. Factoring simplifies the equation and allows us to use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Factoring out sin x gives us:

    sin x (2sin x - 1) = 0

  3. Apply the Zero-Product Property: With the equation factored as sin x (2sin x - 1) = 0, we can apply the zero-product property. This means that either sin x = 0 or 2sin x - 1 = 0. We now have two simpler equations to solve separately. This step is crucial because it breaks down the original equation into more manageable parts. We will find the solutions for each of these equations within the interval [0, 2π).

  4. Solve sin x = 0: The first equation we need to solve is sin x = 0. We need to find all angles x within the interval [0, 2π) where the sine function equals zero. Recall that the sine function corresponds to the y-coordinate on the unit circle. The y-coordinate is zero at the angles 0, π, and 2π. However, since our interval is [0, 2π), we include 0 and π but exclude 2π. Therefore, the solutions for sin x = 0 in the interval [0, 2π) are:

    x = 0, π

  5. Solve 2sin x - 1 = 0: The second equation we need to solve is 2sin x - 1 = 0. To solve for sin x, we first add 1 to both sides and then divide by 2, which gives us:

    sin x = 1/2

    Now we need to find the angles x within the interval [0, 2π) where the sine function equals 1/2. Recall that sin x = 1/2 corresponds to angles in the first and second quadrants where the y-coordinate on the unit circle is 1/2. These angles are π/6 and 5π/6. Therefore, the solutions for sin x = 1/2 in the interval [0, 2π) are:

    x = π/6, 5π/6

  6. Combine the Solutions: Finally, we combine all the solutions we found for both equations. The solutions for the original equation 2sin²x = sin x in the interval [0, 2π) are the solutions from sin x = 0 and sin x = 1/2. These are:

    x = 0, π/6, 5π/6, π

Therefore, the solutions to the trigonometric equation 2sin²x = sin x on the interval [0, 2π) are 0, π/6, 5π/6, and π. These values represent the angles within one full cycle where the equation holds true. Understanding the unit circle and the properties of trigonometric functions is crucial for solving such equations accurately.

Verifying the Solutions

After finding the potential solutions to a trigonometric equation, it's crucial to verify that these solutions are indeed correct. This step ensures that no errors were made during the solving process and that the solutions satisfy the original equation. Verification involves substituting each solution back into the original equation and checking if the equation holds true. This process helps to catch any mistakes and build confidence in the correctness of the solutions.

Method of Verification

  1. Substitute Each Solution: To verify the solutions for the equation 2sin²x = sin x, we substitute each solution (0, π/6, 5π/6, π) back into the original equation. This means replacing x with each of these values and evaluating both sides of the equation.

  2. Evaluate the Equation: For each substitution, we evaluate both the left-hand side (LHS) and the right-hand side (RHS) of the equation. If the LHS equals the RHS for each solution, then the solution is verified. This step involves calculating the sine of each angle and performing the arithmetic operations.

  3. Check for Equality: After evaluating both sides of the equation for each solution, we check if the LHS is equal to the RHS. If they are equal, the solution is valid. If they are not equal, it indicates an error in the solution process, and we need to re-examine our steps.

Verification Process

Let's verify each solution for the equation 2sin²x = sin x:

  1. x = 0:

    • LHS: 2sin²(0) = 2(0)² = 0
    • RHS: sin(0) = 0
    • Since LHS = RHS, x = 0 is a valid solution.

  2. x = π/6:

    • LHS: 2sin²(π/6) = 2(1/2)² = 2(1/4) = 1/2
    • RHS: sin(π/6) = 1/2
    • Since LHS = RHS, x = π/6 is a valid solution.

  3. x = 5π/6:

    • LHS: 2sin²(5π/6) = 2(1/2)² = 2(1/4) = 1/2
    • RHS: sin(5π/6) = 1/2
    • Since LHS = RHS, x = 5π/6 is a valid solution.

  4. x = π:

    • LHS: 2sin²(π) = 2(0)² = 0
    • RHS: sin(π) = 0
    • Since LHS = RHS, x = π is a valid solution.

Conclusion of Verification

After substituting each potential solution into the original equation 2sin²x = sin x and verifying that the left-hand side equals the right-hand side, we can confidently conclude that the solutions are valid. The solutions x = 0, x = π/6, x = 5π/6, and x = π all satisfy the equation within the interval [0, 2π). This verification process is an essential step in solving trigonometric equations, ensuring accuracy and correctness.

Conclusion

In summary, solving trigonometric equations involves a systematic approach that includes rearranging the equation, factoring, applying trigonometric identities, and considering the interval in which solutions are sought. For the equation 2sin²x = sin x, we found the solutions within the interval [0, 2π) to be 0, π/6, 5π/6, and π. These solutions were obtained by factoring the equation, applying the zero-product property, and using our knowledge of the unit circle and trigonometric values. Verifying the solutions by substituting them back into the original equation confirmed their validity.

Trigonometric equations are a fundamental part of trigonometry and have wide-ranging applications in various fields. By mastering the techniques for solving these equations, one can gain a deeper understanding of trigonometric functions and their behavior. The process involves not only algebraic manipulation but also a strong grasp of trigonometric concepts and the unit circle. Consistent practice and a thorough understanding of the underlying principles are key to successfully solving trigonometric equations.

Understanding the solutions to trigonometric equations within a specified interval, such as [0, 2π), is crucial for various applications in mathematics, physics, and engineering. The interval [0, 2π) represents one full cycle of the trigonometric functions, and finding solutions within this interval provides a comprehensive understanding of the equation's behavior. This knowledge is essential for modeling periodic phenomena, such as oscillations, waves, and rotations. By mastering the techniques for solving trigonometric equations, students and professionals can apply these skills to solve real-world problems and gain a deeper appreciation for the beauty and utility of trigonometry.