If `2 sin^2 theta - cos^2 theta , = 2` then find the vlaue of `theta`

Answer

To solve the equation \(2 \sin^2 \theta - \cos^2 \theta = 2\), we will follow these steps:

### Step 1: Use the Pythagorean Identity
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
From this, we can express \(\cos^2 \theta\) in terms of \(\sin^2 \theta\):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]

### Step 2: Substitute \(\cos^2 \theta\) in the Equation
Substituting \(\cos^2 \theta\) in the original equation:
\[ 2 \sin^2 \theta - (1 - \sin^2 \theta) = 2 \]

### Step 3: Simplify the Equation
Now, simplify the equation:
\[ 2 \sin^2 \theta - 1 + \sin^2 \theta = 2 \]
Combine like terms:
\[ 3 \sin^2 \theta - 1 = 2 \]

### Step 4: Isolate \(\sin^2 \theta\)
Add 1 to both sides:
\[ 3 \sin^2 \theta = 3 \]
Now, divide both sides by 3:
\[ \sin^2 \theta = 1 \]

### Step 5: Solve for \(\sin \theta\)
Taking the square root of both sides gives:
\[ \sin \theta = 1 \quad \text{or} \quad \sin \theta = -1 \]

### Step 6: Find the Values of \(\theta\)
The value of \(\theta\) for which \(\sin \theta = 1\) is:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad \text{(for any integer } n\text{)} \]
The value of \(\theta\) for which \(\sin \theta = -1\) is:
\[ \theta = \frac{3\pi}{2} + 2n\pi \quad \text{(for any integer } n\text{)} \]

### Final Answer
Thus, the values of \(\theta\) that satisfy the equation are:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad \text{or} \quad \theta = \frac{3\pi}{2} + 2n\pi \]