Solving Sin(2x) = 0 Solutions In The Interval Π ≤ X ≤ 5π

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In the realm of trigonometry, understanding the behavior of trigonometric functions is crucial. Trigonometric functions, such as sine, cosine, and tangent, exhibit periodic patterns, meaning their values repeat over regular intervals. This periodicity leads to multiple solutions for trigonometric equations within a given range. In this comprehensive exploration, we embark on a journey to determine the number of values of x within the interval π ≤ x ≤ 5π that satisfy the equation sin(2x) = 0. This exploration will delve into the fundamental properties of the sine function, its periodicity, and how these concepts interplay to yield the solutions to the given equation.

Delving into the Sine Function and its Zeros

At the heart of our exploration lies the sine function, a cornerstone of trigonometry. The sine function, denoted as sin(x), maps an angle x to the ratio of the side opposite the angle to the hypotenuse in a right-angled triangle. The sine function's graph oscillates between -1 and 1, exhibiting a wave-like pattern. Understanding the zeros of the sine function is paramount to solving the equation sin(2x) = 0. The zeros of the sine function occur at integer multiples of π, that is, sin(x) = 0 when x = nπ, where n is any integer. This fundamental property stems from the unit circle definition of sine, where the sine corresponds to the y-coordinate of a point on the unit circle. The y-coordinate is zero at angles 0, π, 2π, and so on, leading to the zeros of the sine function at integer multiples of π.

To tackle the equation sin(2x) = 0, we leverage the knowledge of the sine function's zeros. We can rewrite the equation as 2x = nπ, where n is an integer. Solving for x, we obtain x = nπ/2. This equation reveals that the solutions to sin(2x) = 0 occur at half-integer multiples of π. However, our quest is not simply to find all solutions, but to pinpoint those that lie within the specified interval π ≤ x ≤ 5π. This constraint adds a layer of complexity, requiring us to carefully consider which values of n yield solutions within the desired interval. The periodic nature of the sine function means that there will be infinitely many solutions to sin(2x) = 0 if we consider the entire real number line. However, by restricting our attention to the interval π ≤ x ≤ 5π, we effectively narrow down the possibilities and arrive at a finite set of solutions. The challenge now lies in systematically identifying these solutions by considering the half-integer multiples of π and their relationship to the given interval.

Unraveling the Solutions within the Interval π ≤ x ≤ 5π

Now, the crux of the matter lies in determining the integer values of n that produce solutions within the interval π ≤ x ≤ 5π. Substituting x = nπ/2 into the inequality, we get π ≤ nπ/2 ≤ 5π. To isolate n, we multiply all parts of the inequality by 2/π, yielding 2 ≤ n ≤ 10. This inequality unveils the permissible range for n, indicating that n can take on integer values from 2 to 10, inclusive. Each of these values of n corresponds to a unique solution for x within the specified interval. To find these solutions, we systematically substitute each integer value of n from 2 to 10 into the equation x = nπ/2. This process will generate a set of values for x, each of which represents an angle within the interval π ≤ x ≤ 5π where sin(2x) equals zero.

For n = 2, we get x = 2π/2 = π. For n = 3, we get x = 3π/2. Continuing this process, for n = 4, x = 4π/2 = 2π. For n = 5, x = 5π/2. For n = 6, x = 6π/2 = 3π. For n = 7, x = 7π/2. For n = 8, x = 8π/2 = 4π. For n = 9, x = 9π/2. And finally, for n = 10, x = 10π/2 = 5π. We have now generated a comprehensive set of solutions for x: π, 3π/2, 2π, 5π/2, 3π, 7π/2, 4π, 9π/2, and 5π. To determine the total number of solutions, we simply count the elements in this set. We find that there are 9 distinct values of x within the interval π ≤ x ≤ 5π that satisfy the equation sin(2x) = 0. This systematic approach, grounded in the fundamental properties of the sine function and careful consideration of the given interval, has led us to the solution.

The Verdict: Nine Solutions within the Interval

In conclusion, our exploration has revealed that there are precisely 9 values of x within the interval π ≤ x ≤ 5π for which sin(2x) = 0. This determination stemmed from a thorough understanding of the sine function's zeros, its periodic behavior, and the constraints imposed by the specified interval. By systematically identifying the half-integer multiples of π that fall within the interval, we were able to enumerate the solutions and arrive at the final answer. The correct answer is E. 9. This exercise underscores the importance of mastering trigonometric functions and their properties in solving trigonometric equations. The ability to manipulate trigonometric identities, understand periodicity, and apply these concepts within specific intervals is crucial for success in trigonometry and related mathematical fields.