How many solutions does the equation `sec x-1 =(sqrt2-1)tanx` have in the interval `(0,6 pi]`? (A) 6 (B) 5 (C) 10 (D) 9

Options

A. 6
B. 5
C. 10
D. 9

Answer

To solve the equation \( \sec x - 1 = (\sqrt{2} - 1) \tan x \) in the interval \( (0, 6\pi] \), we will 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:
\[ \frac{1}{\cos x} - 1 = (\sqrt{2} - 1) \frac{\sin x}{\cos x} \]

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

### Step 3: Rearrange the equation
Rearranging gives:
\[ 1 - \cos x - (\sqrt{2} - 1) \sin x = 0 \]

### Step 4: Use the identity for \( 1 - \cos x \)
We can use the identity \( 1 - \cos x = 2 \sin^2 \frac{x}{2} \):
\[ 2 \sin^2 \frac{x}{2} - (\sqrt{2} - 1) \sin x = 0 \]
Using the identity \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \):
\[ 2 \sin^2 \frac{x}{2} - (\sqrt{2} - 1)(2 \sin \frac{x}{2} \cos \frac{x}{2}) = 0 \]

### Step 5: Factor out \( 2 \sin \frac{x}{2} \)
Factoring gives:
\[ 2 \sin \frac{x}{2} \left( \sin \frac{x}{2} - (\sqrt{2} - 1) \cos \frac{x}{2} \right) = 0 \]
This results in two cases:
1. \( \sin \frac{x}{2} = 0 \)
2. \( \sin \frac{x}{2} - (\sqrt{2} - 1) \cos \frac{x}{2} = 0 \)

### Step 6: Solve the first case
For \( \sin \frac{x}{2} = 0 \):
\[ \frac{x}{2} = n\pi \quad \Rightarrow \quad x = 2n\pi \]
In the interval \( (0, 6\pi] \), the solutions are:
- \( x = 2\pi \)
- \( x = 4\pi \)
- \( x = 6\pi \)

This gives us **3 solutions**.

### Step 7: Solve the second case
For \( \sin \frac{x}{2} = (\sqrt{2} - 1) \cos \frac{x}{2} \):
Dividing both sides by \( \cos \frac{x}{2} \) (assuming \( \cos \frac{x}{2} \neq 0 \)):
\[ \tan \frac{x}{2} = \sqrt{2} - 1 \]
The general solution for this is:
\[ \frac{x}{2} = \tan^{-1}(\sqrt{2} - 1) + n\pi \]
Thus,
\[ x = 2\tan^{-1}(\sqrt{2} - 1) + 2n\pi \]
Calculating \( \tan^{-1}(\sqrt{2} - 1) \):
Using the fact that \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \):
\[ x = 2 \cdot \frac{\pi}{8} + 2n\pi = \frac{\pi}{4} + 2n\pi \]
In the interval \( (0, 6\pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{4} \)
- For \( n = 1 \): \( x = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} \)
- For \( n = 2 \): \( x = \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} \)

All these values are within \( (0, 6\pi] \).

### Step 8: Count the total solutions
From the first case, we have **3 solutions**: \( 2\pi, 4\pi, 6\pi \).
From the second case, we have **3 solutions**: \( \frac{\pi}{4}, \frac{9\pi}{4}, \frac{17\pi}{4} \).

Thus, the total number of solutions is:
\[ 3 + 3 = 6 \]

### Final Answer
The number of solutions in the interval \( (0, 6\pi] \) is **6**.

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