Problem 141 - Precalculus Essentials

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 difference between the number of solutions of \(\sin x=\frac{1}{2}\) and \(\sin 2 x=\frac{1}{2}\) in the interval \([0,2 \pi)\) is 2. This is because the 2x inside the sin in the second equation doubles the frequency of the function, giving it twice as many solutions in the interval.

Step 1: Find solutions for \(\sin x=\frac{1}{2}\)

Without solving the equation, based on knowledge of the sin function on the interval \([0,2\pi]\), there will be two solutions to \(\sin x=\frac{1}{2}\), which are \(\pi/6\) and \(5\pi/6\).

Step 2: Find solution for \(\sin 2 x=\frac{1}{2}\)

Similarly, the equation \(\sin 2 x=\frac{1}{2}\) will have twice as many solutions as \(\sin x=\frac{1}{2}\) on the interval \([0,2 \pi)\), because the 2x within the sin doubles the frequency of the function. Therefore, there would be four solutions.

Step 3: Calculate the difference

The difference in the number of solutions for the two equations would be the solutions of \(\sin 2 x=\frac{1}{2}\) minus solutions of \(\sin x=\frac{1}{2}\), which is 4 - 2 = 2