Consider the following statements about the expression `sin^(3)theta+2sin^(2)theta+3sintheta`
1. For any `theta in R`, maximum value of this expression is `6` .
2. For any `theta in R`, value of this expression cannot be zero .
Among above statement which is/are true ?

Options

A. only `1`
B. only `2`
C. Both `1` and `2`
D. Neither `1` nor `2`

Answer

To solve the problem, we need to analyze the expression \( \sin^3 \theta + 2\sin^2 \theta + 3\sin \theta \) and evaluate the two statements provided.

### Step 1: Analyze the Expression
The expression is given as:
\[ f(x) = x^3 + 2x^2 + 3x \]
where \( x = \sin \theta \). Since \( \sin \theta \) can take values in the interval \([-1, 1]\), we will analyze \( f(x) \) for \( x \) in this interval.

### Step 2: Find the Maximum Value of the Expression
To find the maximum value of \( f(x) \), we can calculate the derivative and find critical points:
\[ f'(x) = 3x^2 + 4x + 3 \]
Setting the derivative equal to zero to find critical points:
\[ 3x^2 + 4x + 3 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot 3 = 16 - 36 = -20 \]
Since the discriminant is negative, there are no real roots, meaning \( f(x) \) has no critical points in the interval \([-1, 1]\).

### Step 3: Evaluate the Expression at the Endpoints
Now we evaluate \( f(x) \) at the endpoints of the interval:
1. For \( x = 1 \):
\[ f(1) = 1^3 + 2(1^2) + 3(1) = 1 + 2 + 3 = 6 \]
2. For \( x = -1 \):
\[ f(-1) = (-1)^3 + 2(-1)^2 + 3(-1) = -1 + 2 - 3 = -2 \]

### Step 4: Determine the Maximum Value
From the evaluations, we see that the maximum value of \( f(x) \) on the interval \([-1, 1]\) is \( 6 \) at \( x = 1 \). Therefore, the first statement is true.

### Step 5: Check if the Expression Can Be Zero
Next, we check if the expression can be zero:
\[ f(x) = 0 \implies x^3 + 2x^2 + 3x = 0 \]
Factoring out \( x \):
\[ x(x^2 + 2x + 3) = 0 \]
The equation \( x = 0 \) gives \( \sin \theta = 0 \), which occurs at \( \theta = n\pi \) for \( n \in \mathbb{Z} \).

The quadratic \( x^2 + 2x + 3 \) has a discriminant:
\[ D = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \]
Since the discriminant is negative, \( x^2 + 2x + 3 \) has no real roots. Thus, the only solution for \( f(x) = 0 \) is \( x = 0 \).

### Conclusion
1. The maximum value of the expression is indeed \( 6 \), which makes the first statement true.
2. The expression can be zero when \( \sin \theta = 0 \), which makes the second statement false.

Thus, the correct answer is that only the first statement is true.

### Final Answer
Only statement 1 is true.