If `alpha` is a root of `25"cos"^(2) theta+ 5"cos" theta-12 = 0, (pi)/(2) lt alpha lt pi, " then sin"2 alpha` is equal to

Options

A. (a) `(24)/(25)`
B. (b) `-(24)/(25)`
C. (c) `(13)/(18)`
D. (d) `-(13)/(18)`

Answer

To solve the equation \( 25 \cos^2 \theta + 5 \cos \theta - 12 = 0 \) for \( \theta \) in the interval \( \left( \frac{\pi}{2}, \pi \right) \) and find \( \sin 2\alpha \), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \( a x^2 + b x + c = 0 \), where:
- \( a = 25 \)
- \( b = 5 \)
- \( c = -12 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = 5^2 - 4 \cdot 25 \cdot (-12) = 25 + 1200 = 1225 \]

### Step 3: Find the roots using the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ \cos \theta = \frac{-5 \pm \sqrt{1225}}{2 \cdot 25} \]
Calculating \( \sqrt{1225} = 35 \):
\[ \cos \theta = \frac{-5 \pm 35}{50} \]
Calculating the two possible values:
1. \( \cos \theta = \frac{30}{50} = \frac{3}{5} \)
2. \( \cos \theta = \frac{-40}{50} = -\frac{4}{5} \)

### Step 4: Determine the valid root for \( \alpha \)
Since \( \alpha \) is in the interval \( \left( \frac{\pi}{2}, \pi \right) \), we know that \( \cos \alpha \) must be negative. Therefore, we take:
\[ \cos \alpha = -\frac{4}{5} \]

### Step 5: Find \( \sin^2 \alpha \)
Using the Pythagorean identity:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
Substituting \( \cos \alpha \):
\[ \sin^2 \alpha + \left(-\frac{4}{5}\right)^2 = 1 \]
\[ \sin^2 \alpha + \frac{16}{25} = 1 \]
\[ \sin^2 \alpha = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} \]

### Step 6: Find \( \sin 2\alpha \)
Using the double angle formula:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
Substituting the values:
\[ \sin 2\alpha = 2 \cdot \sqrt{\frac{9}{25}} \cdot \left(-\frac{4}{5}\right) \]
Calculating \( \sin \alpha = \sqrt{\frac{9}{25}} = \frac{3}{5} \):
\[ \sin 2\alpha = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = 2 \cdot \frac{3 \cdot (-4)}{25} = \frac{-24}{25} \]

### Final Answer
Thus, the value of \( \sin 2\alpha \) is:
\[ \sin 2\alpha = -\frac{24}{25} \]