The number of points of intersection of 2y = | vedclass

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(D) Given equations are $2y = 1 \implies y = \frac{1}{2}$ and $y = \sin x$. To find the points of intersection, we solve $\sin x = \frac{1}{2}$ for $x

Vedclass · Accepted Answer

$4$

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(D) Given equations are $2y = 1 \implies y = \frac{1}{2}$ and $y = \sin x$. To find the points of intersection, we solve $\sin x = \frac{1}{2}$ for $x \in [-2\pi, 2\pi]$. In the interval $[0, 2\pi]$, $\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. In the interval $[-2\pi, 0]$, $\sin x = \frac{1}{2}$ at $x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$ and $x = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6}$. Thus, the solutions are $x \in \{\frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}, -\frac{11\pi}{6}\}$. There are $4$ such points of intersection.

The number of points of intersection of $2y = 1$ and $y = \sin x$ in the interval $-2\pi \leq x \leq 2\pi$ is:

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