Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) at how many points is \\(\sin \theta = \frac{\sqrt{2}}{2}\\) on the interval \\(0, 2\pi\\)? \\(\circ\\) 4 points \\(\circ\\) 1 point \\(\circ\\) 0 points \\(\circ\\) 2 points

Explanation:

Step1: Recall the unit - circle values

We know that \(\sin\theta=\frac{\sqrt{2}}{2}\) when \(\theta = \frac{\pi}{4}+ 2k\pi\) or \(\theta=\frac{3\pi}{4}+2k\pi\), where \(k\in\mathbb{Z}\) (by the definition of the sine function on the unit circle).

Step2: Find solutions in the interval \([0,2\pi]\)

For \(k = 0\):

• When \(k = 0\) in \(\theta=\frac{\pi}{4}+2k\pi\), we get \(\theta=\frac{\pi}{4}\).
• When \(k = 0\) in \(\theta=\frac{3\pi}{4}+2k\pi\), we get \(\theta=\frac{3\pi}{4}\).

For \(k = 1\):

• If we substitute \(k = 1\) into \(\theta=\frac{\pi}{4}+2k\pi\), \(\theta=\frac{\pi}{4}+2\pi=\frac{9\pi}{4}>2\pi\).
• If we substitute \(k = 1\) into \(\theta=\frac{3\pi}{4}+2k\pi\), \(\theta=\frac{3\pi}{4}+2\pi=\frac{11\pi}{4}>2\pi\).

Answer:

2 points