If `0^(@)ltthetalt90^(@)and2sectheta=3"cosec"^(2)theta` then `theta` is

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(5)`

Answer

To solve the equation \( 2 \sec \theta = 3 \csc^2 \theta \) for \( \theta \) in the interval \( 0^\circ < \theta < 90^\circ \), we will follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
The secant and cosecant functions can be expressed in terms of sine and cosine:
\[ \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta} \]
Thus, we can rewrite the equation as:
\[ 2 \cdot \frac{1}{\cos \theta} = 3 \cdot \left(\frac{1}{\sin \theta}\right)^2 \]
This simplifies to:
\[ \frac{2}{\cos \theta} = \frac{3}{\sin^2 \theta} \]

### Step 2: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 2 \sin^2 \theta = 3 \cos \theta \]

### Step 3: Use the Pythagorean identity
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Therefore, we can express \( \sin^2 \theta \) in terms of \( \cos \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Substituting this into the equation gives:
\[ 2(1 - \cos^2 \theta) = 3 \cos \theta \]
Expanding this results in:
\[ 2 - 2 \cos^2 \theta = 3 \cos \theta \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ 2 \cos^2 \theta + 3 \cos \theta - 2 = 0 \]

### Step 5: Solve the quadratic equation
Let \( x = \cos \theta \). The equation becomes:
\[ 2x^2 + 3x - 2 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = 3, c = -2 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4} \]
This gives us two potential solutions:
\[ x_1 = \frac{2}{4} = \frac{1}{2}, \quad x_2 = \frac{-8}{4} = -2 \]
Since \( \cos \theta \) must be in the range \([-1, 1]\), we discard \( x_2 = -2 \).

### Step 6: Find \( \theta \)
Thus, we have:
\[ \cos \theta = \frac{1}{2} \]
The angle \( \theta \) that satisfies this in the interval \( 0^\circ < \theta < 90^\circ \) is:
\[ \theta = 60^\circ \]

### Final Answer
\[ \theta = 60^\circ \]