Problem 141 - Algebra and Trigonometry

Question

Suppose you are solving equations in the interval \([0,2 \pi)\) Without actually solving equations, what is the difference between the number of solutions of \(\sin x=\frac{1}{2}\) and \(\sin 2 x=\frac{1}{2} ?\) How do you account for this difference?

Step-by-Step Solution

Answer

The number of solutions for \(\sin 2x=\frac{1}{2}\) is two more than the number of solutions for \(\sin x=\frac{1}{2}\) on the interval \([0,2\pi)\). This difference is due to the doubling of the frequency of sinusoid for \(\sin 2x=\frac{1}{2}\), which results in the doubled number of solutions within the same interval.

Step 1: Determine the number of solutions for \(\sin x=\frac{1}{2}\)

By referring to the unit circle, we know that \(\sin x=\frac{1}{2}\) at \(x=\frac{\pi}{6}, \frac{5\pi}{6}\). Therefore, we have two solutions in the interval \([0,2\pi)\).

Step 2: Determine the number of solutions for \(\sin 2x=\frac{1}{2}\)

This equation speeds up the rate at which the sinusoid completes its cycle, so there will be twice as many solutions in the interval \([0,2\pi)\). The solutions are \(x=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{6}, \frac{11\pi}{6}\). Thus, we have four solutions for this equation.

Step 3: Calculate the difference between the number of solutions

The difference of the number of solutions between the two equations is 4 (from \(\sin 2x=\frac{1}{2}\)) - 2 (from \(\sin x=\frac{1}{2}\)) = 2. So, there are two more solutions for \(\sin 2x=\frac{1}{2}\) than \(\sin x=\frac{1}{2}\) on the interval [0, 2pi).