If `-pi//2ltxltpi//2`, and the sum to infinite terms of the series
`cosx+(2)/(3)cosxsin^(2)x+(4)/(9)cosxsin^(4)x+ . . . ` if finite then

Options

A. `x in(-pi//3,pi//3)`
B. `x in(-pi//2,pi//2)`
C. `x in(-pi//4,pi//4)`
D. none of these

Explanation

To solve the problem, we need to analyze the given series and determine the condition under which its sum to infinite terms is finite.

### Step-by-Step Solution:

1. **Identify the Series**:
The series given is:
\[ S = \cos x + \frac{2}{3} \cos x \sin^2 x + \frac{4}{9} \cos x \sin^4 x + \ldots \]

2. **Recognize the Pattern**:
We can observe that the series can be expressed in a form that resembles a geometric progression (GP). The first term \( A_1 \) is \( \cos x \) and the second term \( A_2 \) is \( \frac{2}{3} \cos x \sin^2 x \).

3. **Determine the Common Ratio**:
The common ratio \( r \) of the GP can be calculated as:
\[ r = \frac{A_2}{A_1} = \frac{\frac{2}{3} \cos x \sin^2 x}{\cos x} = \frac{2}{3} \sin^2 x \]

4. **Condition for Finite Sum**:
For the sum of an infinite geometric series to be finite, the common ratio \( r \) must satisfy:
\[ |r| < 1 \]
Therefore, we require:
\[ \left| \frac{2}{3} \sin^2 x \right| < 1 \]

5. **Analyze the Range of \( \sin^2 x \)**:
The function \( \sin^2 x \) varies between 0 and 1 for all \( x \). Thus:
\[ 0 \leq \sin^2 x \leq 1 \]
Multiplying by \( \frac{2}{3} \):
\[ 0 \leq \frac{2}{3} \sin^2 x \leq \frac{2}{3} \]

6. **Check the Condition**:
Since \( \frac{2}{3} \) is less than 1, we have:
\[ 0 \leq \frac{2}{3} \sin^2 x < 1 \]
This condition holds true for all \( x \) in the interval \( -\frac{\pi}{2} < x < \frac{\pi}{2} \).

7. **Conclusion**:
Therefore, the sum of the series is finite for all values of \( x \) in the range:
\[ -\frac{\pi}{2} < x < \frac{\pi}{2} \]

### Final Answer:
The sum to infinite terms of the series is finite for all \( x \) in the interval \( -\frac{\pi}{2} < x < \frac{\pi}{2} \).

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