Solve the equation: `cos^2[pi/4(sinx+sqrt(2)cos^2x)]-tan^2[x+pi/4tan^2x]=1`

Options

A. 1
B. 2
C. 4
D. 8

Answer

To solve the equation

\[ \cos^2\left(\frac{\pi}{4}\left(\sin x + \sqrt{2}\cos^2 x\right)\right) - \tan^2\left(x + \frac{\pi}{4}\tan^2 x\right) = 1, \]

we will follow these steps:

### Step 1: Simplify the Equation
We start by rewriting the equation:

\[ \cos^2\left(\frac{\pi}{4}\left(\sin x + \sqrt{2}\cos^2 x\right)\right) = 1 + \tan^2\left(x + \frac{\pi}{4}\tan^2 x\right). \]

Using the identity \( \cos^2 A = 1 - \sin^2 A \) and \( 1 + \tan^2 B = \sec^2 B \), we can rewrite the left-hand side as:

\[ 1 - \sin^2\left(\frac{\pi}{4}\left(\sin x + \sqrt{2}\cos^2 x\right)\right) = \sec^2\left(x + \frac{\pi}{4}\tan^2 x\right). \]

### Step 2: Set Up the Equation
This leads us to:

\[ \sin^2\left(\frac{\pi}{4}\left(\sin x + \sqrt{2}\cos^2 x\right)\right) = 1 - \sec^2\left(x + \frac{\pi}{4}\tan^2 x\right). \]

Since \( \sec^2 B \) is always greater than or equal to 1, the right-hand side is non-positive, which means the left-hand side must also equal zero:

\[ \sin^2\left(\frac{\pi}{4}\left(\sin x + \sqrt{2}\cos^2 x\right)\right) = 0. \]

### Step 3: Solve for \( x \)
The sine function equals zero at integer multiples of \( \pi \):

\[ \frac{\pi}{4}\left(\sin x + \sqrt{2}\cos^2 x\right) = n\pi \quad \text{for } n \in \mathbb{Z}. \]

This simplifies to:

\[ \sin x + \sqrt{2}\cos^2 x = 4n. \]

### Step 4: Substitute and Rearrange
Rearranging gives us:

\[ \sin x = 4n - \sqrt{2}\cos^2 x. \]

Using the identity \( \cos^2 x = 1 - \sin^2 x \):

\[ \sin x = 4n - \sqrt{2}(1 - \sin^2 x). \]

### Step 5: Form a Quadratic Equation
This leads to:

\[ \sqrt{2}\sin^2 x + \sin x + (4n - \sqrt{2}) = 0. \]

### Step 6: Solve the Quadratic
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = \sqrt{2}, b = 1, c = 4n - \sqrt{2} \):

\[ \sin x = \frac{-1 \pm \sqrt{1 - 4\sqrt{2}(4n - \sqrt{2})}}{2\sqrt{2}}. \]

### Step 7: Find Solutions
We will need to check the discriminant to ensure it is non-negative for real solutions.

### Step 8: General Solutions
For each valid \( n \), we can find corresponding \( x \) values using:

\[ x = \arcsin\left(\text{value}\right) + 2k\pi \quad \text{and} \quad x = \pi - \arcsin\left(\text{value}\right) + 2k\pi \quad (k \in \mathbb{Z}). \]