Consider the function f(x)=sin(x) in the interval x=[π/4,7π/4]. The number and location(s) of the local minima of this function are? This is MCQ of a competitive exam(GATE), Answer is (d) given by GATE , and from other sources ,explanation is (b) somewhere and (d) somewhere , I am going with (b) as minimum at $270$, I have drawn graph . but it produces two pictures , I confused , it asked Local minima of a trivial $\sin(x)$ function and explanation here, but didn't get obviously , finally I want to explanation . 
Problem is :

Consider the function $f(x)=\sin(x)$ in the interval $x=[\pi /4,7 \pi /4]$. The
  number and location(s) of the local minima of this function are
(A) One, at $\pi /2$ 
(B) One, at $3 \pi /2$ 
(C) Two, at $ \pi /2$ and $3 \pi /2$
(D) Two, at $\pi /4$ and $3 \pi /2$

 A: The local minima is at $x=\frac{3\pi}{2}$
This is very obvious from the graph of $f(x)=\sin{x}$
On a second look at the graph below, I believe $x=\frac{\pi}{4}$ is also a local minimum. This is because it is lesser than all other values within its locality. 
Thus we have two local minima: $x=\frac{\pi}{4}, \frac{3\pi}{2}$

A: Why not use a definition...?
According to Wikipedia Maxima and minima — Definition:

If the domain $X$ is a metric space then $f$ is said to have a local (or relative) [...] minimum point at $x^∗$ if [there exists some $ε > 0$ such that] $f(x^∗) ≤ f(x)$ for all $x$ in $X$ within distance $ε$ of $x^∗$

Of course $\pi/4$ fits the definition criteria: for any (sufficiently small) neighborhood of $\pi/4$, sine at that point is smaller than for any other point in the neghborhood. So $\pi/4$ is the local minimum point.
A: 
Edit: Starting from the initial point $x=\frac{\pi}{4}$, we see that $f(x)=\sin x$ clearly has a maxima at $x=\frac{\pi}{2}$ & $x=\frac{\pi}{4}$ is in the given  domain $\left[\frac{\pi}{4}, \frac{7\pi}{4}\right]$  of the function & it is minimum in the locality of $x$ in $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$  Hence $x=\frac{\pi}{4}$ is also considered as a point of minima. 

And $f(x)=\sin x$ clearly has a minima at $x=\frac{3\pi}{2}$ which is mimimum in the locality of $x$ $\left[\frac{3\pi}{2}, \frac{7\pi}{4}\right]$ 
Thus, in the given interval $\left[\frac{\pi}{4}, \frac{7\pi}{4}\right]$ there are two points of maxima one at $x=\frac{\pi}{4}$ & other at $x=\frac{3\pi}{2}$ 
Hence, the option (D) is correct. 
