If `max{5sin theta+3sin(theta-alpha)}=7` then the set of possible vaues of `alpha` is `theta in R`

Options

A. `{x:x=2 n pi pm (pi)/(3), n in I}`
B. `{x:x=2 n pi pm (2pi)/(3), n in I}`
C. `[(pi)/(3),(2pi)/(3)]`
D. None of these

Answer

To solve the problem, we need to find the set of possible values of \( \alpha \) given that the maximum value of the expression \( 5 \sin \theta + 3 \sin(\theta - \alpha) \) is equal to 7.

### Step 1: Rewrite the expression
We start with the expression:
\[ 5 \sin \theta + 3 \sin(\theta - \alpha) \]
Using the sine subtraction formula, we can rewrite \( \sin(\theta - \alpha) \) as:
\[ \sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha \]
Substituting this back into the expression gives:
\[ 5 \sin \theta + 3 (\sin \theta \cos \alpha - \cos \theta \sin \alpha) = (5 + 3 \cos \alpha) \sin \theta - 3 \sin \alpha \cos \theta \]

### Step 2: Factor out the sine and cosine terms
We can factor out \( \sin \theta \) and \( \cos \theta \):
\[ (5 + 3 \cos \alpha) \sin \theta - 3 \sin \alpha \cos \theta \]
This can be expressed in the form:
\[ A \sin \theta + B \cos \theta \]
where \( A = 5 + 3 \cos \alpha \) and \( B = -3 \sin \alpha \).

### Step 3: Find the maximum value of the expression
The maximum value of \( A \sin \theta + B \cos \theta \) is given by:
\[ \sqrt{A^2 + B^2} \]
Substituting for \( A \) and \( B \):
\[ \sqrt{(5 + 3 \cos \alpha)^2 + (-3 \sin \alpha)^2} \]
This simplifies to:
\[ \sqrt{(5 + 3 \cos \alpha)^2 + 9 \sin^2 \alpha} \]

### Step 4: Set the maximum value equal to 7
According to the problem, this maximum value is equal to 7:
\[ \sqrt{(5 + 3 \cos \alpha)^2 + 9 \sin^2 \alpha} = 7 \]
Squaring both sides gives:
\[ (5 + 3 \cos \alpha)^2 + 9 \sin^2 \alpha = 49 \]

### Step 5: Expand and simplify
Expanding the left side:
\[ (5 + 3 \cos \alpha)^2 = 25 + 30 \cos \alpha + 9 \cos^2 \alpha \]
Thus, we have:
\[ 25 + 30 \cos \alpha + 9 \cos^2 \alpha + 9 \sin^2 \alpha = 49 \]
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ 25 + 30 \cos \alpha + 9(1) = 49 \]
This simplifies to:
\[ 34 + 30 \cos \alpha = 49 \]
Subtracting 34 from both sides gives:
\[ 30 \cos \alpha = 15 \]
Dividing by 30:
\[ \cos \alpha = \frac{1}{2} \]

### Step 6: Find values of \( \alpha \)
The solutions for \( \cos \alpha = \frac{1}{2} \) are:
\[ \alpha = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Final Answer
The set of possible values of \( \alpha \) is:
\[ \alpha = 2n\pi + \frac{\pi}{3} \quad \text{or} \quad \alpha = 2n\pi - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]