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 will follow these steps:

### Step 1: Rewrite the equation using the Pythagorean identity
We know that \( \sin^2 \beta + \cos^2 \beta = 1 \). Therefore, we can express \( \cos^2 \beta \) in terms of \( \sin^2 \beta \):
\[ \cos^2 \beta = 1 - \sin^2 \beta \]
Substituting this into the original equation gives:
\[ 2 \sin^2 \beta - (1 - \sin^2 \beta) = 2 \]

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

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

### Step 4: Solve for \( \sin \beta \)
Taking the square root of both sides gives:
\[ \sin \beta = \pm 1 \]

### Step 5: Determine the values of \( \beta \)
The sine function equals 1 at \( \beta = 90^\circ \) and equals -1 at \( \beta = 270^\circ \). Thus, the possible values of \( \beta \) are:
\[ \beta = 90^\circ \quad \text{or} \quad \beta = 270^\circ \]

### Final Answer
The values of \( \beta \) are \( 90^\circ \) and \( 270^\circ \).

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