on the interval ( 0 leq x leq 2pi ), what are the ( x )-intercepts of ( y=sin x )?\n( 0, \frac{pi}{2} ), and…

on the interval ( 0 leq x leq 2pi ), what are the ( x )-intercepts of ( y=sin x )?\n( 0, \frac{pi}{2} ), and ( \frac{3pi}{2} )\n( 0, pi ), and ( 2pi )\n( \frac{pi}{2} ) and ( \frac{3pi}{2} ) only\n( pi ) and ( 2pi ) only

on the interval ( 0 leq x leq 2pi ), what are the ( x )-intercepts of ( y=sin x )?\n( 0, \frac{pi}{2} ), and ( \frac{3pi}{2} )\n( 0, pi ), and ( 2pi )\n( \frac{pi}{2} ) and ( \frac{3pi}{2} ) only\n( pi ) and ( 2pi ) only

Answer

Explanation:

Step1: Recall the definition of x - intercept

The x - intercepts of the function (y = f(x)) are the values of (x) for which (y = 0). So, we need to solve the equation (\sin x=0) for (x\in[0, 2\pi]).

Step2: Solve the trigonometric equation

We know that the general solution of the equation (\sin x = k) is given by (x=n\pi+(- 1)^{n}\arcsin k), where (n\in\mathbb{Z}). For (k = 0), (\arcsin(0)=0), and the equation (\sin x=0) has solutions (x = n\pi), (n\in\mathbb{Z}). When (n = 0), (x=0); when (n = 1), (x=\pi); when (n = 2), (x = 2\pi) and these values (x = 0,\pi,2\pi) lie in the interval (0\leq x\leq2\pi).

Answer:

0, (\pi), and (2\pi) (the second option)