91Ó°ÊÓ

In trigonometric equations, it's important to find the general solution to get a complete picture of all possible answers. The general solution takes into account the periodic nature of trigonometric functions and provides a formula that captures all possible solutions. For the sine function, the general solution for any angle \(\theta\) where \(\sin(\theta) = y\) is written as \(\theta = \text{angle} + 2k\pi\). Here, \(k\) is any integer, and this term represents the infinite number of times the function repeats its values in one complete cycle of \(2\pi\) radians.

For example, in the provided exercise, solving \(\sin(3\pi x) = 1\) involves finding all angles where the sine is equal to 1. According to the general solution, this happens at angles \(\theta = \frac{\pi}{2} + 2k\pi\). Substituting \(\theta = 3\pi x\) into this, we get \(3\pi x = \frac{\pi}{2} + 2k\pi\). Solving for \(x\), we obtain \(x = \frac{1}{6} + \frac{2k}{3}\), where \(k\) can be any integer.

The sine function is crucial in trigonometry. It maps an angle to the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. The sine function is periodic with a period of \(2\pi\), meaning that the function repeats its values every \(2\pi\) units. The range of the sine function is \([-1, 1]\), and it achieves its maximum value of 1 and minimum value of -1 at specific points.

In the given problem, the goal is to find when \(\sin(3\pi x) = 1\). The sine function equals 1 at angles where \(\theta = \frac{\pi}{2} + 2k\pi\), representing the peak points of the sine wave within each cycle of \(2\pi\). Understanding this behavior is essential for solving trigonometric equations accurately and comprehensively.

In trigonometric equations, integer values are used to represent all possible repeats of the function's cycles. The variable \(k\) is introduced to account for these repeats. By allowing \(k\) to be any integer, we effectively capture the infinite nature of trigonometric functions.

For example, in our solution \(\sin(3\pi x) = 1\), the integer \(k\) ensures we find all values for \(x\) that satisfy the equation. Substituting multiple \(k\) values into \(x = \frac{1}{6} + \frac{2k}{3}\) gives us a series of solutions, revealing the cyclic nature of the problem and ensuring we don't miss any possible solutions. Thus, the integer values play a fundamental role in general solutions for trigonometric equations, helping to frame all the circumstances in which the original condition is satisfied.