Q.The equality $\sin A + \sin 2A + \sin 3A = 3$ holds for some real value of $A$.

The key idea here is understanding the range of the sine function.

For the equality $\sin A + \sin 2A + \sin 3A = 3$ to hold, each term must individually attain its maximum possible value. We know that for any real angle $x$, $-1 \le \sin x \le 1$.

1. For the sum of three sine terms to be 3, each term must be exactly 1. That is, $\sin A = 1$, $\sin 2A = 1$, and $\sin 3A = 1$.
2. If $\sin A = 1$, then $A$ must be of the form $2n\pi + \frac{\pi}{2}$ for some integer $n$.
3. Substitute this form of $A$ into $\sin 2A$: $2A = 2\left(2n\pi + \frac{\pi}{2}\right) = 4n\pi + \pi$. …

The problem asks us to consider the statement: "The equality $\sin A + \sin 2A + \sin 3A = 3$ holds for some real value of $A$." To determine if this statement is true or false, we need to investigate if there exists any real number $A$ that satisfies this equation.

The core concept here is the range of the sine function. We know that for any real angle $x$, the value of $\sin x$ is always between $-1$ and $1$, inclusive. That is, $-1 \le \sin x \le 1$. This fundamental property will be key to analyzing the given sum.

Let's break down the analysis step by step.

1. Understanding the Range of the Sine Function

   For any real number $x$, the sine function $\sin x$ has a maximum value of $1$ and a minimum value of $-1$.

   This means:

   $\sin A \le 1$

   $\sin 2A \le 1$

   $\sin 3A \le 1$

2. Deducing Conditions for the Sum to be $3$

   We are given the equation $\sin A + \sin 2A + \sin 3A = 3$.

   Since each term on the left side can be at most $1$, the maximum possible value for the sum $\sin A + \sin 2A + \sin 3A$ is $1 + 1 + 1 = 3$.

   For the sum to actually reach this maximum value of $3$, each individual term must simultaneously attain its maximum value of $1$.

   Therefore, for the equality to hold, we must have:

   $\sin A = 1$

   $\sin 2A = 1$

   $\sin 3A = 1$

3. Solving the First Condition: $\sin A = 1$

   Let's find the general solution for $A$ when $\sin A = 1$.

   The general solution for $\sin x = 1$ is $x = 2n\pi + \frac{\pi}{2}$, where $n$ is an integer.

   The general solution for $\sin x = 1$ is $x = 2n\pi + \frac{\pi}{2}$, where $n \in \mathbb{Z}$.

   So, for $\sin A = 1$, we must have:

   $A = 2n\pi + \frac{\pi}{2}$ for some integer $n$.

4. Checking Consistency with the Second Condition: $\sin 2A = 1$

   Now, we substitute the expression for $A$ from step 3 into the second condition, $\sin 2A = 1$.

   $2A = 2 \left( 2n\pi + \frac{\pi}{2} \right)$

   $2A = 4n\pi + \pi$

   Now, let's evaluate $\sin(2A)$ using this expression:

   $\sin(2A) = \sin(4n\pi + \pi)$ …