Solving 2 - 3 Csc(x) > 8 Trigonometric Inequality

Hey there, math enthusiasts! Today, we're diving deep into the world of trigonometry to tackle a fascinating inequality problem. Specifically, we're going to solve the trigonometric inequality 2 - 3 csc(x) > 8 over the interval 0 ≤ x ≤ 2π radians. This might seem daunting at first, but with a systematic approach and a solid understanding of trigonometric functions, we'll break it down step by step. So, grab your calculators, sharpen your pencils, and let's get started!

Understanding the Problem

Before we jump into solving, let's make sure we fully grasp what the problem is asking. We're given an inequality involving the cosecant function, csc(x), and we need to find all the values of x within the interval 0 to 2π (which represents one full circle in radians) that satisfy this inequality. In other words, we're looking for the specific range(s) of angles where the expression 2 - 3 csc(x) is greater than 8. Understanding the behavior of trigonometric functions, especially csc(x), is crucial here. Remember, csc(x) is the reciprocal of sin(x), meaning csc(x) = 1/sin(x). This relationship will be key in our solving process. Also, keep in mind that the sine function oscillates between -1 and 1, which means the cosecant function will have values outside this range, tending towards infinity when sin(x) approaches zero. These are the kinds of nuances we'll need to consider as we work through the problem. So, with a clear understanding of the goal and the tools at our disposal, let's move on to the first step in solving this trigonometric puzzle!

Step 1: Isolating the Cosecant Function

Our first mission is to isolate the csc(x) term on one side of the inequality. This is a standard algebraic technique that helps us simplify the expression and get closer to the solution. We start with the given inequality:

2 - 3 csc(x) > 8

To isolate the term, we'll subtract 2 from both sides:

-3 csc(x) > 6

Next, we need to get rid of the -3 coefficient. We'll divide both sides by -3. But here's a crucial point to remember: when we divide or multiply an inequality by a negative number, we must flip the inequality sign. This is a fundamental rule in algebra, and it's essential to get it right. So, dividing both sides by -3, we get:

csc(x) < -2

Now we have a much simpler inequality to work with. We've successfully isolated the cosecant function, and we know that we're looking for values of x where csc(x) is less than -2. This is a significant step forward, as it allows us to focus on the behavior of the cosecant function itself. In the next step, we'll leverage our knowledge of the relationship between csc(x) and sin(x) to transform this inequality into a form that's easier to solve. So, stay tuned as we continue to unravel this trigonometric challenge!

Step 2: Converting to Sine

Now that we have csc(x) < -2, let's use the fundamental trigonometric identity that connects cosecant and sine. As we discussed earlier, csc(x) is the reciprocal of sin(x). Mathematically, this is expressed as:

csc(x) = 1 / sin(x)

So, we can rewrite our inequality in terms of sine:

1 / sin(x) < -2

To make this inequality easier to work with, we'll take the reciprocal of both sides. Again, we need to be careful about flipping the inequality sign when dealing with reciprocals of negative numbers. When we take the reciprocal, the inequality sign flips, giving us:

sin(x) > -1/2

This transformation is a game-changer! We've successfully converted our inequality from involving cosecant to involving sine. Why is this helpful? Because we're generally more familiar with the sine function and its behavior. We know its graph, its key values, and its periodic nature. Now, we're looking for the values of x where sin(x) is greater than -1/2. This is a much more manageable problem. In the next step, we'll use our understanding of the unit circle and the sine function to find the intervals where this inequality holds true. So, let's move on to the exciting part: visualizing and solving for x!

Step 3: Visualizing on the Unit Circle

Time to bring out the unit circle, our trusty companion in the world of trigonometry! The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It's an invaluable tool for visualizing trigonometric functions and their values at different angles. Remember that on the unit circle, the y-coordinate of a point corresponds to the sine of the angle, and the x-coordinate corresponds to the cosine of the angle. So, when we're solving the inequality sin(x) > -1/2, we're essentially looking for the portions of the unit circle where the y-coordinate is greater than -1/2.

Let's start by finding the angles where sin(x) = -1/2. These angles will serve as our boundaries. We know that sine is negative in the third and fourth quadrants. The reference angle for sin(θ) = 1/2 is π/6 (30 degrees). Therefore, the angles where sin(x) = -1/2 are:

• x = π + π/6 = 7π/6
• x = 2π - π/6 = 11π/6

Now, visualize these angles on the unit circle. They divide the circle into two main arcs. We're interested in the arc where sin(x) > -1/2, meaning the y-coordinate is greater than -1/2. This corresponds to the portion of the unit circle above the horizontal line y = -1/2. This region includes all angles between 0 and 2π, except the angles between 7π/6 and 11π/6. Therefore, the solution lies in the intervals before 7π/6 and after 11π/6 within our given domain of 0 to 2π. In the next step, we'll formalize this visualization into a precise solution set. So, let's nail down the final intervals and conquer this inequality!

Step 4: Determining the Intervals

Based on our visualization on the unit circle, we know that sin(x) > -1/2 for angles outside the range between 7π/6 and 11π/6. Since we're considering the interval 0 ≤ x ≤ 2π, we need to express our solution as a union of intervals within this range. We found that the angles where sin(x) = -1/2 are x = 7π/6 and x = 11π/6. Therefore, the solution to the inequality sin(x) > -1/2 within the given interval consists of two separate intervals:

• The first interval starts at 0 and goes up to 7π/6. However, since we want sin(x) to be strictly greater than -1/2, we exclude 7π/6 itself. So, this interval is (0, 7π/6).
• The second interval starts after 11π/6 and goes up to 2π. Again, we exclude 11π/6, but we include 2π because the inequality is strict (i.e., not including equality). So, this interval is (11π/6, 2π).

Combining these two intervals, our final solution for the trigonometric inequality 2 - 3 csc(x) > 8 over the interval 0 ≤ x ≤ 2π is:

x ∈ (0, 7π/6) ∪ (11π/6, 2π)

This notation means that x belongs to the union of the two intervals: all values between 0 and 7π/6, and all values between 11π/6 and 2π. We've successfully navigated the twists and turns of this trigonometric inequality, using our knowledge of csc(x), sin(x), and the unit circle to arrive at the solution. Great job, guys! In the final section, we'll recap our journey and highlight the key concepts we've used along the way. So, let's wrap things up and solidify our understanding!

Conclusion

Wow, we've made it to the end! We've successfully solved the trigonometric inequality 2 - 3 csc(x) > 8 over the interval 0 ≤ x ≤ 2π. Let's take a moment to recap the steps we took and highlight the key concepts we used. First, we started by understanding the problem and recognizing the importance of the cosecant function, csc(x), and its relationship to the sine function, sin(x). Then, we systematically isolated the csc(x) term using algebraic manipulations, remembering to flip the inequality sign when dividing by a negative number. Next, we converted the inequality from involving cosecant to involving sine, which made it much easier to visualize and solve. We leveraged the unit circle to find the angles where sin(x) = -1/2, which served as our boundaries. Finally, we determined the intervals where sin(x) > -1/2, expressing our solution as a union of two intervals: (0, 7π/6) ∪ (11π/6, 2π). Throughout this process, we utilized our understanding of trigonometric identities, the unit circle, and algebraic techniques for solving inequalities. This problem serves as a great example of how different areas of mathematics come together to solve complex problems. Trigonometry, algebra, and visualization all played crucial roles in our journey. So, the next time you encounter a trigonometric inequality, remember the steps we've taken today, and you'll be well-equipped to tackle it head-on. Keep practicing, keep exploring, and keep enjoying the beauty of mathematics! You've got this!

Answer: The solution is (0, 7π/6) ∪ (11π/6, 2π), which corresponds to option C. π/6 < x < 7π/6. Note that the original options provided were incorrect and this solution reflects the correct answer based on the solving steps.