If `|1-(|sin x |)/(1+ sin x )| ge (2)/(3)`, then sin x lies in

Options

A. `(-oo,(-1)/(2)] cup [(1)/(2),oo)`
B. `(-(1)/(2),(1)/(2))`
C. `[-1,(-5)/(8)] cup [-(1)/(4),(1)/(2)]`
D. None of the above

Answer

To solve the inequality \( \left| 1 - \frac{|\sin x|}{1 + \sin x} \right| \geq \frac{2}{3} \), we will break it down into cases based on the value of \( \sin x \).

### Step 1: Case 1 - \( \sin x \geq 0 \)

In this case, \( |\sin x| = \sin x \). Therefore, we rewrite the inequality as:

\[ \left| 1 - \frac{\sin x}{1 + \sin x} \right| \geq \frac{2}{3} \]

This simplifies to:

\[ 1 - \frac{\sin x}{1 + \sin x} \geq \frac{2}{3} \]

### Step 2: Solve the inequality

To eliminate the absolute value, we can directly solve:

\[ 1 - \frac{\sin x}{1 + \sin x} \geq \frac{2}{3} \]

Subtracting 1 from both sides gives:

\[ -\frac{\sin x}{1 + \sin x} \geq -\frac{1}{3} \]

Multiplying through by -1 (which reverses the inequality):

\[ \frac{\sin x}{1 + \sin x} \leq \frac{1}{3} \]

### Step 3: Cross-multiply

Cross-multiplying gives:

\[ 3\sin x \leq 1 + \sin x \]

Rearranging this results in:

\[ 2\sin x \leq 1 \implies \sin x \leq \frac{1}{2} \]

### Step 4: Combine with the condition

Since we assumed \( \sin x \geq 0 \), we have:

\[ 0 \leq \sin x \leq \frac{1}{2} \]

### Step 5: Case 2 - \( \sin x < 0 \)

In this case, \( |\sin x| = -\sin x \). The inequality becomes:

\[ \left| 1 + \frac{\sin x}{1 + \sin x} \right| \geq \frac{2}{3} \]

This simplifies to:

\[ 1 + \frac{\sin x}{1 + \sin x} \leq -\frac{2}{3} \]

### Step 6: Solve this inequality

This leads to:

\[ \frac{\sin x}{1 + \sin x} \leq -\frac{5}{3} \]

This inequality will not hold since \( \sin x \) is negative and \( 1 + \sin x \) is also positive (as \( \sin x < 0 \) implies \( 1 + \sin x < 1 \)). Thus, we will consider another approach.

### Step 7: Break into subcases

1. **Subcase 2a**: \( 1 + 2\sin x \geq 0 \)

Here, we can write:

\[ 1 + 2\sin x \geq \frac{2}{3}(1 + \sin x) \]

Rearranging gives:

\[ 2\sin x - \frac{2}{3}\sin x \geq \frac{2}{3} - 1 \]

Simplifying yields:

\[ \frac{4}{3}\sin x \geq -\frac{1}{3} \implies \sin x \geq -\frac{1}{4} \]

Thus, we have:

\[ -\frac{1}{4} \leq \sin x < 0 \]

2. **Subcase 2b**: \( 1 + 2\sin x < 0 \)

Here, we have:

\[ -1 - 2\sin x \geq \frac{2}{3}(1 + \sin x) \]

Rearranging gives:

\[ -3 - 6\sin x \geq 2 + 2\sin x \]

Thus:

\[ -8\sin x \leq 5 \implies \sin x \leq -\frac{5}{8} \]

### Step 8: Combine the results

Now we combine the results from both cases:

1. From Case 1: \( 0 \leq \sin x \leq \frac{1}{2} \)
2. From Case 2a: \( -\frac{1}{4} \leq \sin x < 0 \)
3. From Case 2b: \( \sin x \leq -\frac{5}{8} \)

### Final Result

Thus, the solution set for \( \sin x \) is:

\[ \sin x \in \left[-1, -\frac{5}{8}\right) \cup \left[0, \frac{1}{2}\right] \]