Solving -2cos²x + √3 Cos X + 3 > 0 Trigonometric Inequality

This article delves into the solution of the trigonometric inequality -2cos²x + √3 cos x + 3 > 0. We will explore the steps involved in solving this inequality, providing a comprehensive explanation that will be helpful for students and anyone interested in trigonometry. We will break down the problem, making it easier to understand and follow the solution process. Mastering trigonometric inequalities is a crucial step in advanced mathematics and has applications in various fields such as physics and engineering.

1. Understanding Trigonometric Inequalities

Before diving into the specifics of the given inequality, let's establish a solid understanding of trigonometric inequalities in general. Trigonometric inequalities involve trigonometric functions such as sine, cosine, tangent, and their reciprocals. Unlike trigonometric equations that seek specific values where the expressions are equal, trigonometric inequalities seek ranges of values where the trigonometric expression satisfies a given inequality (>, <, ≥, ≤). Understanding trigonometric inequalities requires a strong foundation in trigonometric identities, unit circle concepts, and algebraic manipulation techniques. When solving these inequalities, it's crucial to consider the periodic nature of trigonometric functions and their respective ranges. The cosine function, for instance, oscillates between -1 and 1, a fact that plays a vital role in determining the solution set. We also need to be adept at using trigonometric identities to simplify complex expressions and transform them into a more manageable form. For example, identities like sin²x + cos²x = 1 or double-angle formulas can be invaluable tools. Visualizing the graphs of trigonometric functions is also extremely helpful. The graph of cos(x) clearly shows its periodic nature and its values between -1 and 1, which aids in identifying the intervals that satisfy a given inequality. Additionally, the unit circle provides a visual representation of trigonometric values for different angles, making it easier to understand the sign and magnitude of sine, cosine, and tangent functions. A thorough understanding of the unit circle is essential for determining the solutions of trigonometric inequalities. By combining algebraic techniques, trigonometric identities, and graphical analysis, we can effectively solve a wide range of trigonometric inequalities. This foundational knowledge forms the backbone for tackling more complex problems and applications in calculus and other areas of mathematics. Furthermore, proficiency in solving trigonometric inequalities is not just an academic exercise; it has practical applications in fields like physics, engineering, and computer graphics, where oscillatory phenomena and periodic behavior are frequently encountered.

2. Transforming the Inequality

The initial step in solving the inequality -2cos²x + √3 cos x + 3 > 0 involves transforming it into a more manageable form. This transformation typically involves treating the trigonometric function, in this case, cos x, as a variable. Let's substitute y = cos x. This substitution converts the trigonometric inequality into a quadratic inequality: -2y² + √3y + 3 > 0. Now, we have a standard quadratic inequality that we can solve using algebraic techniques. The next step is to multiply the entire inequality by -1 to make the leading coefficient positive, which simplifies the factoring process. Remember that multiplying an inequality by a negative number reverses the inequality sign. So, we get: 2y² - √3y - 3 < 0. This quadratic inequality is now in a form that is easier to factor or solve using the quadratic formula. The goal here is to find the values of y that satisfy this inequality. Once we find the range of y values, we can then substitute back cos x for y and find the corresponding values of x. This substitution technique is a common strategy in solving trigonometric equations and inequalities, allowing us to leverage our knowledge of algebraic techniques to solve trigonometric problems. It's a powerful method that simplifies complex expressions and makes them more accessible. By transforming the trigonometric inequality into a quadratic inequality, we can apply well-established algebraic methods to find the solution set. This approach highlights the interconnectedness of algebra and trigonometry and demonstrates how algebraic techniques can be used to solve trigonometric problems effectively. The ability to make such transformations is a key skill in mathematical problem-solving, allowing us to tackle complex problems by breaking them down into simpler, more manageable parts.

3. Solving the Quadratic Inequality

Now that we have the quadratic inequality 2y² - √3y - 3 < 0, our next step is to solve it. Solving the quadratic inequality can be done by factoring or by using the quadratic formula to find the roots of the corresponding quadratic equation. Let's first attempt to factor the quadratic expression. However, it might not be immediately obvious how to factor it. In such cases, the quadratic formula provides a reliable alternative. The quadratic formula states that for an equation of the form ay² + by + c = 0, the roots are given by: y = (-b ± √(b² - 4ac)) / (2a). Applying the quadratic formula to our equation 2y² - √3y - 3 = 0, we have a = 2, b = -√3, and c = -3. Plugging these values into the formula, we get: y = (√3 ± √((-√3)² - 4 * 2 * (-3))) / (2 * 2). Simplifying this, we have: y = (√3 ± √(3 + 24)) / 4 = (√3 ± √27) / 4. Since √27 = 3√3, the roots are: y = (√3 ± 3√3) / 4. This gives us two roots: y₁ = (√3 + 3√3) / 4 = 4√3 / 4 = √3 and y₂ = (√3 - 3√3) / 4 = -2√3 / 4 = -√3 / 2. Now we know the roots of the quadratic equation, and we can use them to determine the intervals where the quadratic inequality 2y² - √3y - 3 < 0 holds true. Since the parabola opens upwards (because the coefficient of y² is positive), the inequality is satisfied between the roots. Therefore, the solution to the quadratic inequality is -√3 / 2 < y < √3. This means that the values of y that satisfy the inequality lie between -√3 / 2 and √3. This range of values will be crucial when we substitute back cos x for y to find the solution to the original trigonometric inequality. Understanding how to solve quadratic inequalities is fundamental in various areas of mathematics, and this step demonstrates its application in the context of trigonometric inequalities.

4. Substituting Back and Finding the General Solution

After solving the quadratic inequality, we found that -√3 / 2 < y < √3. Now, we need to substitute back cos x for y to return to our original trigonometric inequality. Substituting back and finding the general solution means we now have -√3 / 2 < cos x < √3. However, we know that the range of the cosine function is -1 ≤ cos x ≤ 1. Therefore, we need to consider the intersection of the interval (-√3 / 2, √3) with the range of the cosine function, which is [-1, 1]. Since √3 ≈ 1.732, the upper bound √3 is outside the range of the cosine function. Thus, the inequality simplifies to -√3 / 2 < cos x ≤ 1. Now, we need to find the angles x for which cos x lies within this interval. First, let's find the angles where cos x = -√3 / 2. The reference angle for cos⁻¹(√3 / 2) is π / 6 radians or 30 degrees. Cosine is negative in the second and third quadrants. Therefore, the angles in the interval [0, 2π) where cos x = -√3 / 2 are x = π - π / 6 = 5π / 6 and x = π + π / 6 = 7π / 6. Since we want cos x to be greater than -√3 / 2, we are looking for the intervals where the cosine function is above the value -√3 / 2. This occurs between the angles 5π / 6 and 7π / 6. Therefore, the solution to the inequality in the interval [0, 2π) is 5π / 6 < x < 7π / 6. However, since the cosine function is periodic with a period of 2π, we need to find the general solution. To do this, we add integer multiples of 2π to our solution. The general solution is given by: 5π / 6 + 2πk < x < 7π / 6 + 2πk, where k is an integer. This general solution represents all the angles x for which the original trigonometric inequality holds true. It's important to remember the periodic nature of trigonometric functions when solving inequalities, as it ensures that we capture all possible solutions. The general solution provides a complete picture of the solution set, covering all possible values of x that satisfy the given inequality.

5. Verifying the Solution

Once we have obtained a solution, it is crucial to verify its correctness. Verifying the solution ensures that we haven't made any errors in our calculations and that our solution indeed satisfies the original inequality. To verify our solution 5π / 6 + 2πk < x < 7π / 6 + 2πk, where k is an integer, we can test values within and outside this interval. Let's first consider k = 0, which gives us the interval 5π / 6 < x < 7π / 6. Let's choose a value within this interval, say x = π (which is between 5π / 6 and 7π / 6). Plugging x = π into the original inequality -2cos²x + √3 cos x + 3 > 0, we get: -2cos²(π) + √3 cos(π) + 3 = -2(-1)² + √3(-1) + 3 = -2 - √3 + 3 = 1 - √3. Since √3 ≈ 1.732, 1 - √3 is approximately -0.732, which is not greater than 0. This indicates a potential issue with our solution. Let's re-examine our steps. We made an error in determining the interval where cos x > -√3 / 2. The correct interval should be from 0 to 5π / 6 and from 7π / 6 to 2π. So, the solution in the interval [0, 2π) is 0 ≤ x < 5π / 6 and 7π / 6 < x ≤ 2π. Therefore, the general solution should be: 2πk ≤ x < 5π / 6 + 2πk and 7π / 6 + 2πk < x ≤ 2π + 2πk, where k is an integer. Now let's verify this corrected solution. Choose x = π / 2 (which is in the interval 0 ≤ x < 5π / 6). Plugging x = π / 2 into the original inequality, we get: -2cos²(π / 2) + √3 cos(π / 2) + 3 = -2(0)² + √3(0) + 3 = 3, which is greater than 0. This confirms that the interval 0 ≤ x < 5π / 6 is part of the solution. Now, choose x = 3π / 2 (which is in the interval 7π / 6 < x ≤ 2π). Plugging x = 3π / 2 into the original inequality, we get: -2cos²(3π / 2) + √3 cos(3π / 2) + 3 = -2(0)² + √3(0) + 3 = 3, which is also greater than 0. This confirms that the interval 7π / 6 < x ≤ 2π is also part of the solution. The importance of verification cannot be overstated. It is a critical step in the problem-solving process that helps us identify and correct errors, ensuring that our final answer is accurate. By testing values within and outside the solution intervals, we can gain confidence in the correctness of our solution.

Conclusion

In this article, we have walked through the process of solving the trigonometric inequality -2cos²x + √3 cos x + 3 > 0. We began by transforming the inequality into a quadratic inequality, solved the quadratic inequality, substituted back to the trigonometric form, found the general solution, and finally, verified our solution. Solving trigonometric inequalities involves a combination of algebraic techniques, trigonometric identities, and an understanding of the periodic nature of trigonometric functions. It also highlights the importance of careful verification to ensure the accuracy of our results. The skills and concepts covered in this article are fundamental in trigonometry and have applications in various fields of mathematics, science, and engineering. Mastering these skills is essential for anyone pursuing advanced studies in these areas. The ability to solve trigonometric inequalities is not only a valuable mathematical skill but also a powerful tool for understanding and modeling real-world phenomena that exhibit periodic behavior. By following the steps outlined in this article and practicing similar problems, one can develop a strong foundation in solving trigonometric inequalities.