Understanding the Range of a Trigonometric Function

The problem asks for the range of the function defined by \(f(x) = \pi + \sin^2 x\).

To find the range of this function, we need to understand the range of the core trigonometric part, which is \(\sin^2 x\).

We know that the range of the sine function, \(\sin x\), is \([-1, 1]\). This means that for any real number \(x\), the value of \(\sin x\) is between -1 and 1, inclusive:

\(-1 \le \sin x \le 1\)

Now let's consider \(\sin^2 x\). When we square a number between -1 and 1, the result will be a number between 0 and 1. For example, \((-1)^2 = 1\), \(0^2 = 0\), and \(1^2 = 1\). Any value between -1 and 1, when squared, will fall in the range \([0, 1]\).

So, the range of \(\sin^2 x\) is \([0, 1]\):

\(0 \le \sin^2 x \le 1\)

The given function is \(f(x) = \pi + \sin^2 x\). This means we are adding the constant value \(\pi\) to \(\sin^2 x\). To find the range of \(f(x)\), we add \(\pi\) to each part of the inequality for \(\sin^2 x\):

\(0 + \pi \le \pi + \sin^2 x \le 1 + \pi\)

\(\pi \le f(x) \le \pi + 1\)

Therefore, the range of the function \(f(x) = \pi + \sin^2 x\) is the interval \([\pi, \pi + 1]\).

This range starts at \(\pi\) when \(\sin^2 x\) is at its minimum value (0) and goes up to \(\pi + 1\) when \(\sin^2 x\) is at its maximum value (1).

Let's summarise the steps:

1. Identify the basic trigonometric function and its range (\(\sin x\) has range \([-1, 1]\)).
2. Determine the range of the squared trigonometric function (\(\sin^2 x\) has range \([0, 1]\)).
3. Add the constant term (\(\pi\)) to the range of \(\sin^2 x\) to find the range of \(f(x)\).

The range of \(f(x) = \pi + \sin^2 x\) is \([\pi, \pi + 1]\).