which formula gives the zeros of ( y=sin(x) )?\n( kpi ) for any positive integer ( k )\n( kpi ) for any…

which formula gives the zeros of ( y=sin(x) )?\n( kpi ) for any positive integer ( k )\n( kpi ) for any integer ( k )\n( \frac{kpi}{2} ) for any positive integer ( k )\n( \frac{kpi}{2} ) for any integer ( k )

which formula gives the zeros of ( y=sin(x) )?\n( kpi ) for any positive integer ( k )\n( kpi ) for any integer ( k )\n( \frac{kpi}{2} ) for any positive integer ( k )\n( \frac{kpi}{2} ) for any integer ( k )

Answer

Explanation:

Step1: Recall the zero - points of the sine function

The sine function (y = \sin(x)) has zeros when (x = k\pi), where (k\in\mathbb{Z}). This is because (\sin(k\pi)=0) for all integers (k). When (k = 0), (\sin(0)=0); when (k = 1), (\sin(\pi)=0); when (k=- 1), (\sin(-\pi)=0) and so on.

Step2: Analyze each option

  • Option 1: (k\pi) for any positive integer (k) misses the case when (k = 0) or (k) is negative. For example, when (k = 0), (\sin(0)=0) and when (k=-1), (\sin(-\pi)=0).
  • Option 2: (k\pi) for any integer (k) is correct since (\sin(k\pi)=0) for all (k\in\mathbb{Z}).
  • Option 3: (\frac{k\pi}{2}) for any positive integer (k). When (k = 1), (\sin(\frac{\pi}{2}) = 1\neq0).
  • Option 4: (\frac{k\pi}{2}) for any integer (k). When (k = 1), (\sin(\frac{\pi}{2})=1\neq0).

Answer:

B. (k\pi) for any integer (k)