For $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$, the number of points of intersection of curves $y = \cos x$ and $y = \sin 3x$ is
Step-by-Step Solution
Key Concept: Convert trigonometric equation to standard form using complementary angle relationships and solve using periodicity.
Given $y = \cos x$ and $y = \sin 3x$, we need to find intersections. Setting $\cos x = \cos(\frac{\pi}{2} - 3x)$, we get $-4x = 2n\pi + \frac{\pi}{2}$ or $-2x = 2n\pi - \frac{\pi}{2}$. Solving these: $x = \frac{n\pi}{2} + \frac{\pi}{8}$ and $x = (4k+1)\frac{\pi}{2}$. For $n = 0$: $x = \frac{\pi}{8}$; for $n = -1$: $x = -\frac{3\pi}{8}$. In $[0,2\pi]$, we get $x = \frac{\pi}{8}$ and $x = \frac{7\pi}{8}$, giving 3 solutions total in the given interval.
Correct Answer: 3