Find the number of distinct solution of ` sec x + tan x =sqrt(3)` , where ` 0 le x le 3pi`

Answer

To solve the equation \( \sec x + \tan x = \sqrt{3} \) for the interval \( 0 \leq x \leq 3\pi \), we can follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Substituting these into the equation gives:
\[ \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sqrt{3} \]
Combining the fractions:
\[ \frac{1 + \sin x}{\cos x} = \sqrt{3} \]

### Step 2: Clear the fraction
Multiply both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ 1 + \sin x = \sqrt{3} \cos x \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ \sqrt{3} \cos x - \sin x = 1 \]

### Step 4: Use the substitution method
We can express this in the form \( R \cos(x - \alpha) = 1 \). Here, we need to find \( R \) and \( \alpha \):
- Let \( R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2 \)
- To find \( \alpha \), we use:
\[ \tan \alpha = \frac{-1}{\sqrt{3}} \implies \alpha = -\frac{\pi}{6} \]

### Step 5: Rewrite the equation
Thus, we can rewrite the equation as:
\[ 2 \cos(x + \frac{\pi}{6}) = 1 \]
Dividing both sides by 2 gives:
\[ \cos(x + \frac{\pi}{6}) = \frac{1}{2} \]

### Step 6: Solve for \( x \)
The general solutions for \( \cos \theta = \frac{1}{2} \) are:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2n\pi \]
Substituting back for \( \theta = x + \frac{\pi}{6} \):
1. \( x + \frac{\pi}{6} = \frac{\pi}{3} + 2n\pi \)
2. \( x + \frac{\pi}{6} = -\frac{\pi}{3} + 2n\pi \)

### Step 7: Solve each case
**Case 1:**
\[ x + \frac{\pi}{6} = \frac{\pi}{3} + 2n\pi \implies x = \frac{\pi}{3} - \frac{\pi}{6} + 2n\pi = \frac{\pi}{6} + 2n\pi \]

**Case 2:**
\[ x + \frac{\pi}{6} = -\frac{\pi}{3} + 2n\pi \implies x = -\frac{\pi}{3} - \frac{\pi}{6} + 2n\pi = -\frac{2\pi}{6} - \frac{\pi}{6} + 2n\pi = -\frac{3\pi}{6} + 2n\pi = -\frac{\pi}{2} + 2n\pi \]

### Step 8: Find distinct solutions in the interval \( [0, 3\pi] \)
1. For \( n = 0 \):
- From Case 1: \( x = \frac{\pi}{6} \)
- From Case 2: \( x = -\frac{\pi}{2} \) (not valid)

2. For \( n = 1 \):
- From Case 1: \( x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6} \)
- From Case 2: \( x = -\frac{\pi}{2} + 2\pi = \frac{3\pi}{2} \)

3. For \( n = 2 \):
- From Case 1: \( x = \frac{\pi}{6} + 4\pi \) (not valid)
- From Case 2: \( x = -\frac{\pi}{2} + 4\pi \) (not valid)

### Step 9: Collect valid solutions
The valid solutions in the interval \( [0, 3\pi] \) are:
- \( x = \frac{\pi}{6} \)
- \( x = \frac{3\pi}{2} \)
- \( x = \frac{13\pi}{6} \)

### Conclusion
Thus, the number of distinct solutions is **3**.