The number of solutions of the equation `"tan" theta + "sec" theta = 2 "cos" theta` lying the interval `[0, 2pi]` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( \tan \theta + \sec \theta = 2 \cos \theta \) for the number of solutions in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan \theta + \sec \theta = 2 \cos \theta \]
We can rewrite \( \tan \theta \) and \( \sec \theta \) in terms of sine and cosine:
\[ \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = 2 \cos \theta \]

### Step 2: Combine the left side
Taking the common denominator on the left side:
\[ \frac{\sin \theta + 1}{\cos \theta} = 2 \cos \theta \]

### Step 3: Cross-multiply
Cross-multiplying gives:
\[ \sin \theta + 1 = 2 \cos^2 \theta \]

### Step 4: Use the Pythagorean identity
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ \sin \theta + 1 = 2(1 - \sin^2 \theta) \]
Expanding the right side:
\[ \sin \theta + 1 = 2 - 2 \sin^2 \theta \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 2 \sin^2 \theta + \sin \theta - 1 = 0 \]

### Step 6: Solve the quadratic equation
This is a quadratic equation in terms of \( \sin \theta \). We can use the quadratic formula:
\[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 1, c = -1 \):
\[ \sin \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \sin \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]

### Step 7: Find the values of \( \sin \theta \)
Calculating the two possible values:
1. \( \sin \theta = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \)
2. \( \sin \theta = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \)

### Step 8: Determine the angles corresponding to these sine values
1. For \( \sin \theta = \frac{1}{2} \):
- The angles are \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).

2. For \( \sin \theta = -1 \):
- The angle is \( \theta = \frac{3\pi}{2} \).

### Step 9: Count the solutions in the interval \( [0, 2\pi] \)
The solutions in the interval \( [0, 2\pi] \) are:
- \( \frac{\pi}{6} \)
- \( \frac{5\pi}{6} \)
- \( \frac{3\pi}{2} \)

Thus, the total number of solutions is **3**.

### Final Answer
The number of solutions of the equation \( \tan \theta + \sec \theta = 2 \cos \theta \) lying in the interval \( [0, 2\pi] \) is **3**.

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