Solving 2cos(θ) + √3 = 0 Find All Solutions In Radians

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Introduction

In this article, we will delve into the process of finding all solutions to the trigonometric equation 2cos(θ) + √3 = 0. Trigonometric equations are fundamental in mathematics and have wide applications in various fields such as physics, engineering, and computer graphics. Solving these equations involves finding the angles (θ) that satisfy the given condition. Our primary focus will be on expressing the solutions in radians, using π as the unit, and accounting for the periodic nature of trigonometric functions. We'll explore the underlying principles, step-by-step methodology, and the significance of representing solutions in a general form that encompasses all possible angles.

Understanding Trigonometric Equations

Trigonometric equations, at their core, are equations that involve trigonometric functions such as sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). The solutions to these equations are the angles that, when plugged into the trigonometric functions, make the equation true. Unlike algebraic equations which may have a finite number of solutions, trigonometric equations often have infinitely many solutions due to the periodic nature of trigonometric functions. This periodicity means that the functions repeat their values at regular intervals. For example, the sine and cosine functions repeat every 2π radians, and the tangent function repeats every π radians. Therefore, when solving trigonometric equations, it is essential to find not just one or two solutions, but a general form that represents all possible solutions. This is typically done by adding integer multiples of the period to the principal solutions.

Solving the Equation 2cos(θ) + √3 = 0

To solve the equation 2cos(θ) + √3 = 0, we will follow a systematic approach. First, we isolate the cosine function on one side of the equation. Then, we find the principal angles that satisfy the equation within one period (usually 0 to 2π). Finally, we express the general solutions by adding integer multiples of the period to these principal angles. This ensures that we capture all possible solutions due to the periodic nature of the cosine function.

Step-by-Step Solution

  1. Isolate the cosine function: Our initial equation is 2cos(θ) + √3 = 0. To isolate cos(θ), we subtract √3 from both sides, giving us 2cos(θ) = -√3. Then, we divide both sides by 2 to obtain cos(θ) = -√3/2. This form is now easier to work with as we can directly relate it to known cosine values.

  2. Identify reference angles: Now we need to find the angles θ for which cos(θ) = -√3/2. Recall that cosine is the x-coordinate on the unit circle. The value -√3/2 is a standard cosine value associated with angles in the second and third quadrants. The reference angle (the angle in the first quadrant with the same cosine value magnitude) for cos(θ) = √3/2 is π/6 radians (30 degrees). However, we are looking for angles where the cosine is negative, which occurs in the second and third quadrants.

  3. Find principal solutions: In the second quadrant, the angle θ is given by π - π/6 = 5π/6. In the third quadrant, the angle θ is given by π + π/6 = 7π/6. These are the two principal solutions within the interval [0, 2π). They represent the angles where the cosine function equals -√3/2 within one full rotation around the unit circle.

  4. Express general solutions: Since the cosine function has a period of 2π, we can add integer multiples of 2π to our principal solutions to find all possible solutions. The general solution can be expressed as: θ = 5π/6 + 2kπ or θ = 7π/6 + 2kπ, where k is an integer (k ∈ Z). This notation represents an infinite set of solutions, as each integer value of k corresponds to a different angle that satisfies the original equation. The term 2kπ accounts for the periodicity, ensuring we capture all angles coterminal with 5π/6 and 7π/6.

Verification of Solutions

To ensure the correctness of our solutions, we can substitute the general solutions back into the original equation and verify that they hold true. Let’s check both solutions:

  1. For θ = 5π/6 + 2kπ: 2cos(5π/6 + 2kπ) + √3 = 2(-√3/2) + √3 = -√3 + √3 = 0. This verifies the solution for all integer values of k.

  2. For θ = 7π/6 + 2kπ: 2cos(7π/6 + 2kπ) + √3 = 2(-√3/2) + √3 = -√3 + √3 = 0. This also verifies the solution for all integer values of k.

The verification confirms that both general solutions are correct and satisfy the given trigonometric equation.

Importance of General Solutions

In many practical applications, it's crucial to identify all possible solutions to a trigonometric equation. For instance, when modeling periodic phenomena such as oscillations or wave behavior, the general solution provides a complete picture of the system's state at any given time. The general solutions, expressed as θ = 5π/6 + 2kπ and θ = 7π/6 + 2kπ, are comprehensive because they include every angle that satisfies the equation 2cos(θ) + √3 = 0. This contrasts with simply finding a few solutions within a specific interval, which might not be sufficient for certain applications. Understanding and expressing general solutions allows for a more thorough analysis and prediction of the behavior of systems described by trigonometric functions.

Conclusion

In conclusion, we have successfully found all solutions to the trigonometric equation 2cos(θ) + √3 = 0. The solutions are expressed in radians in terms of π as θ = 5π/6 + 2kπ and θ = 7π/6 + 2kπ, where k is an integer. We achieved this by isolating the cosine function, identifying reference angles, finding principal solutions, and expressing general solutions to account for the periodic nature of the cosine function. This process underscores the importance of understanding trigonometric functions and their properties, particularly their periodicity, in solving trigonometric equations. By expressing solutions in a general form, we ensure that we capture all possible angles that satisfy the equation, which is crucial for various applications in mathematics, science, and engineering. Solving trigonometric equations like this one forms the foundation for more advanced topics in trigonometry and calculus, making it a vital skill for anyone studying these fields. Remember, the key to solving trigonometric equations lies in understanding the unit circle, the behavior of trigonometric functions in different quadrants, and the concept of periodicity.