The solution of `sqrt(5-2sinx) ge 6 sinx-1` is

Options

A. `[pi(12n-7)//6,pi(12n+7)//6] (n in Z)`
B. `[pi(12n-7)//6,pi(12n+1)//6] (n in Z)`
C. `[pi(2n-7)//6,pi(2n+1)//6] (n in Z)`
D. `[pi(12n-7)//3,pi(12n+1)//3] (n in Z)`

Answer

To solve the inequality \( \sqrt{5 - 2\sin x} \geq 6\sin x - 1 \), we will follow these steps:

### Step 1: Square both sides
First, we square both sides of the inequality to eliminate the square root. However, we must remember that squaring both sides can introduce extraneous solutions, so we will need to check our solutions later.

\[ 5 - 2\sin x \geq (6\sin x - 1)^2 \]

### Step 2: Expand the right side
Next, we expand the right side:

\[ (6\sin x - 1)^2 = 36\sin^2 x - 12\sin x + 1 \]

So the inequality becomes:

\[ 5 - 2\sin x \geq 36\sin^2 x - 12\sin x + 1 \]

### Step 3: Rearrange the inequality
Now, we rearrange the inequality to bring all terms to one side:

\[ 0 \geq 36\sin^2 x - 12\sin x + 1 + 2\sin x - 5 \]

This simplifies to:

\[ 0 \geq 36\sin^2 x - 10\sin x - 4 \]

### Step 4: Rearranging further
Rearranging gives us:

\[ 36\sin^2 x - 10\sin x - 4 \leq 0 \]

### Step 5: Solve the quadratic inequality
Now we will solve the quadratic equation \( 36\sin^2 x - 10\sin x - 4 = 0 \) using the quadratic formula:

\[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 36 \), \( b = -10 \), and \( c = -4 \):

\[ \sin x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 36 \cdot (-4)}}{2 \cdot 36} \]

Calculating the discriminant:

\[ \sqrt{100 + 576} = \sqrt{676} = 26 \]

So we have:

\[ \sin x = \frac{10 \pm 26}{72} \]

Calculating the two possible values:

1. \( \sin x = \frac{36}{72} = \frac{1}{2} \)
2. \( \sin x = \frac{-16}{72} = -\frac{2}{9} \)

### Step 6: Analyze the intervals
Now we have the critical points \( \sin x = \frac{1}{2} \) and \( \sin x = -\frac{2}{9} \).

The sine function is periodic, so we need to find the intervals where the quadratic \( 36\sin^2 x - 10\sin x - 4 \leq 0 \) holds true.

### Step 7: Test intervals
We can test the intervals determined by the roots \( -\frac{2}{9} \) and \( \frac{1}{2} \):

1. For \( \sin x < -\frac{2}{9} \)
2. For \( -\frac{2}{9} < \sin x < \frac{1}{2} \)
3. For \( \sin x > \frac{1}{2} \)

Using a sign chart or test points, we find that the quadratic is less than or equal to zero in the intervals:

- \( \sin x \in [-\frac{2}{9}, \frac{1}{2}] \)

### Step 8: Find corresponding x values
Now we need to find the values of \( x \) for which \( \sin x = -\frac{2}{9} \) and \( \sin x = \frac{1}{2} \):

1. \( \sin x = \frac{1}{2} \) gives \( x = \frac{\pi}{6}, \frac{5\pi}{6} \)
2. \( \sin x = -\frac{2}{9} \) gives two solutions in the range \( [0, 2\pi] \).

### Step 9: Combine intervals
The final solution will be the union of the intervals where \( \sin x \) is between these values.

### Final Answer
The solution set is:

\[ x \in \left[0, \frac{\pi}{6}\right] \cup \left[\frac{5\pi}{6}, \frac{13\pi}{6}\right] \]