`"cosec"^(2)theta(cos^(2)theta-3cos theta+2)ge1`, If `theta` belongs to

Options

A. `(0, (pi)/(3))`
B. `((pi)/(2), pi)`
C. `((pi)/(3), (pi)/(2))`
D. `(0, (pi)/(4))`

Answer

To solve the inequality \( \csc^2 \theta (\cos^2 \theta - 3 \cos \theta + 2) \geq 1 \), we will follow these steps:

### Step 1: Rewrite the inequality in terms of sine and cosine
We know that \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \). Thus, we can rewrite the inequality as:

\[ \frac{1}{\sin^2 \theta} (\cos^2 \theta - 3 \cos \theta + 2) \geq 1 \]

### Step 2: Multiply both sides by \( \sin^2 \theta \)
Since \( \sin^2 \theta > 0 \) for \( \theta \) not equal to \( n\pi \) (where \( n \) is an integer), we can multiply both sides by \( \sin^2 \theta \):

\[ \cos^2 \theta - 3 \cos \theta + 2 \geq \sin^2 \theta \]

### Step 3: Substitute \( \sin^2 \theta \)
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can substitute this into the inequality:

\[ \cos^2 \theta - 3 \cos \theta + 2 \geq 1 - \cos^2 \theta \]

### Step 4: Rearrange the inequality
Rearranging gives us:

\[ \cos^2 \theta + \cos^2 \theta - 3 \cos \theta + 2 - 1 \geq 0 \]

This simplifies to:

\[ 2 \cos^2 \theta - 3 \cos \theta + 1 \geq 0 \]

### Step 5: Factor the quadratic
Now, we will factor the quadratic \( 2 \cos^2 \theta - 3 \cos \theta + 1 \):

\[ (2 \cos \theta - 1)(\cos \theta - 1) \geq 0 \]

### Step 6: Find the critical points
Setting each factor to zero gives us the critical points:

1. \( 2 \cos \theta - 1 = 0 \) → \( \cos \theta = \frac{1}{2} \) → \( \theta = \frac{\pi}{3} \)
2. \( \cos \theta - 1 = 0 \) → \( \cos \theta = 1 \) → \( \theta = 0 \)

### Step 7: Analyze the intervals
We need to analyze the sign of the product \( (2 \cos \theta - 1)(\cos \theta - 1) \) in the intervals defined by the critical points \( 0, \frac{\pi}{3}, \) and \( \pi \):

- For \( \theta < 0 \): Not applicable since \( \theta \) must be in the range of angles where cosine is defined.
- For \( 0 < \theta < \frac{\pi}{3} \): Choose \( \theta = \frac{\pi}{6} \) → \( (2 \cdot \frac{\sqrt{3}}{2} - 1)(\frac{\sqrt{3}}{2} - 1) < 0 \)
- For \( \frac{\pi}{3} < \theta < 1 \): Choose \( \theta = \frac{\pi}{2} \) → \( (2 \cdot 0 - 1)(0 - 1) > 0 \)
- For \( \theta > 1 \): Choose \( \theta = \pi \) → \( (2 \cdot -1 - 1)(-1 - 1) < 0 \)

### Step 8: Write the solution
The solution to the inequality is:

\[ \theta \in \left[0, \frac{\pi}{3}\right] \cup \left[\frac{\pi}{3}, \pi\right] \]