Solve `secx-1=(sqrt(2)-1)tanx`

Answer

To solve the equation \( \sec x - 1 = (\sqrt{2} - 1) \tan x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sec x - 1 = (\sqrt{2} - 1) \tan x \]

### Step 2: Square both sides
Next, we square both sides to eliminate the tangent:
\[ (\sec x - 1)^2 = ((\sqrt{2} - 1) \tan x)^2 \]

### Step 3: Expand both sides
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we expand both sides:
\[ \sec^2 x - 2 \sec x + 1 = (\sqrt{2} - 1)^2 \tan^2 x \]
Calculating \( (\sqrt{2} - 1)^2 \):
\[ (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]
Thus, we rewrite the equation as:
\[ \sec^2 x - 2 \sec x + 1 = (3 - 2\sqrt{2}) \tan^2 x \]

### Step 4: Substitute \( \tan^2 x \)
Using the identity \( \tan^2 x = \sec^2 x - 1 \), we substitute:
\[ \sec^2 x - 2 \sec x + 1 = (3 - 2\sqrt{2})(\sec^2 x - 1) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side gives:
\[ \sec^2 x - 2 \sec x + 1 = (3 - 2\sqrt{2}) \sec^2 x - (3 - 2\sqrt{2}) \]

### Step 6: Rearrange the equation
Rearranging terms, we get:
\[ \sec^2 x - (3 - 2\sqrt{2}) \sec^2 x - 2 \sec x + 1 + (3 - 2\sqrt{2}) = 0 \]
This simplifies to:
\[ (1 - (3 - 2\sqrt{2})) \sec^2 x - 2 \sec x + (4 - 2\sqrt{2}) = 0 \]
\[ (2\sqrt{2} - 2) \sec^2 x - 2 \sec x + (4 - 2\sqrt{2}) = 0 \]

### Step 7: Factor out common terms
Dividing through by 2, we have:
\[ (\sqrt{2} - 1) \sec^2 x - \sec x + (2 - \sqrt{2}) = 0 \]

### Step 8: Solve the quadratic equation
Let \( y = \sec x \). The equation becomes:
\[ (\sqrt{2} - 1) y^2 - y + (2 - \sqrt{2}) = 0 \]
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{1 \pm \sqrt{(-1)^2 - 4(\sqrt{2} - 1)(2 - \sqrt{2})}}{2(\sqrt{2} - 1)} \]

### Step 9: Calculate the discriminant
Calculating the discriminant:
\[ 1 - 4(\sqrt{2} - 1)(2 - \sqrt{2}) = 1 - 4(2\sqrt{2} - 2 - 2 + \sqrt{2}) = 1 - 4(3 - 3\sqrt{2}) \]
This simplifies to:
\[ 1 - 12 + 12\sqrt{2} = 12\sqrt{2} - 11 \]

### Step 10: Find values of \( y \)
Substituting back into the quadratic formula gives us values for \( y \), which is \( \sec x \).

### Step 11: Determine \( x \)
Once we find \( \sec x \), we can find \( x \) using:
\[ \sec x = \frac{1}{\cos x} \]
Thus, \( x = \cos^{-1}(\frac{1}{\sec x}) \).

### Step 12: General solution
The general solution will be:
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]