If 2 `sin^2beta-cos^2beta=2`, then `beta` is:

Options

A. `0^@`
B. `90^@`
C. `45^@`
D. `30^@`

Answer

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

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

### Step 2: Substitute \(\cos^2 \beta\) in the equation
Now, we can substitute \(\cos^2 \beta\) in the original equation:
\[ 2 \sin^2 \beta - (1 - \sin^2 \beta) = 2 \]
This simplifies to:
\[ 2 \sin^2 \beta - 1 + \sin^2 \beta = 2 \]

### Step 3: Combine like terms
Combine the terms involving \(\sin^2 \beta\):
\[ 3 \sin^2 \beta - 1 = 2 \]

### Step 4: Solve for \(\sin^2 \beta\)
Now, add 1 to both sides:
\[ 3 \sin^2 \beta = 3 \]
Next, divide by 3:
\[ \sin^2 \beta = 1 \]

### Step 5: Find \(\beta\)
Taking the square root of both sides gives us:
\[ \sin \beta = 1 \quad \text{or} \quad \sin \beta = -1 \]
This implies:
\[ \beta = 90^\circ \quad \text{or} \quad \beta = -90^\circ \]

### Conclusion
Thus, the possible values for \(\beta\) are \(90^\circ\) and \(-90^\circ\). Since \(90^\circ\) is a common angle in trigonometry, we can conclude that:
\[ \beta = 90^\circ \]