Here is a simple solution to the given problem with clear explanations:

(i) The Image Set of $f : \mathbb{R} \to \mathbb{R}$ where $f(x) = \sin x$:

1. The function $f(x) = \sin x$ is defined for all real numbers $x$, and the sine function is periodic with a range between $-1$ and $1$.

2. This means that no matter what value $x$ takes in $\mathbb{R}$, the output of $\sin x$ will always lie in the interval: $[-1, 1].$

3. Therefore, the image set of $f$, which is the set of all possible outputs of $f(x)$, is: $\text{Image}(f) = [-1, 1].$

4. To determine whether $f$ is surjective (onto):

   • A function $f : \mathbb{R} \to \mathbb{R}$ is surjective if every element in the codomain ($\mathbb{R}$) has at least one pre-image in $\mathbb{R}$.
   • However, the image set of $f$ is not equal to the codomain $\mathbb{R}$, because $f(x)$ cannot take values outside $[-1, 1]$.

Conclusion: The function $f(x) = \sin x$ is not surjective because it does not cover all real numbers in the codomain $\mathbb{R}$.

(ii) Solving $f(x) = \frac{1}{2}$ and Checking Injectivity:

1. To find all $x \in \mathbb{R}$ such that $f(x) = \sin x = \frac{1}{2}$:

   • The sine function equals $\frac{1}{2}$ at specific angles: $x = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad x = \pi - \frac{\pi}{6} + 2n\pi \quad \text{for} \, n \in \mathbb{Z}.$
   • Simplifying these values: $x = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2n\pi \quad \text{where} \, n \in \mathbb{Z}.$

2. These solutions indicate that there are infinitely many values of $x$ that satisfy $f(x) = \frac{1}{2}$. For example:

   • $x = \frac{\pi}{6}$, $x = \frac{5\pi}{6}$, $x = \frac{\pi}{6} + 2\pi$, $x = \frac{5\pi}{6} + 2\pi$, and so on.

3. To determine whether $f$ is injective (one-to-one):

   • A function is injective if each output corresponds to exactly one input.
   • However, for $f(x) = \frac{1}{2}$, there are multiple $x$-values (as shown above) that map to the same output.

Conclusion: The function $f(x) = \sin x$ is not injective because it maps multiple inputs to the same output.

Final Summary:

• (i) The image set of $f$ is $[-1, 1]$, and $f$ is not surjective.
• (ii) $f(x) = \sin x$ is not injective because multiple inputs produce the same output.

Do you want further clarification or details on any step? Let me know!

Here are 5 related questions to explore further:

1. What is the range of the cosine function $f(x) = \cos x$?
2. How do we check injectivity and surjectivity of a quadratic function?
3. What is the period of the sine function and why is it important?
4. How can we graph the function $f(x) = \sin x$ and interpret its range?
5. What are the differences between injective, surjective, and bijective functions?

Tip: A function is injective if it maps different inputs to different outputs and surjective if it covers the entire codomain. For sine and cosine functions, their periodicity often breaks injectivity.