Solve for `theta(0^(@) lt theta lt 90^(@))` :
`2sin^(2)theta-2costheta=1/2`

Answer

To solve the equation \( 2\sin^2\theta - 2\cos\theta = \frac{1}{2} \) for \( \theta \) in the interval \( 0^\circ < \theta < 90^\circ \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\sin^2\theta - 2\cos\theta = \frac{1}{2} \]

### Step 2: Use the Pythagorean identity
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \), we can express \( \sin^2\theta \) in terms of \( \cos\theta \):
\[ \sin^2\theta = 1 - \cos^2\theta \]
Substituting this into the equation gives:
\[ 2(1 - \cos^2\theta) - 2\cos\theta = \frac{1}{2} \]

### Step 3: Simplify the equation
Expanding and simplifying the equation:
\[ 2 - 2\cos^2\theta - 2\cos\theta = \frac{1}{2} \]
Now, multiply through by 2 to eliminate the fraction:
\[ 4 - 4\cos^2\theta - 4\cos\theta = 1 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 4\cos^2\theta + 4\cos\theta - 3 = 0 \]

### Step 5: Factor the quadratic equation
Let \( A = \cos\theta \). The equation becomes:
\[ 4A^2 + 4A - 3 = 0 \]
To factor this, we look for two numbers that multiply to \( -12 \) (the product of \( 4 \times -3 \)) and add to \( 4 \). The numbers \( 6 \) and \( -2 \) work:
\[ 4A^2 + 6A - 2A - 3 = 0 \]
Grouping gives:
\[ 2A(2A + 3) - 1(2A + 3) = 0 \]
Factoring out \( (2A + 3) \):
\[ (2A + 3)(2A - 1) = 0 \]

### Step 6: Solve for \( A \)
Setting each factor to zero gives:
1. \( 2A + 3 = 0 \) → \( A = -\frac{3}{2} \) (not valid since \( A \) must be between -1 and 1)
2. \( 2A - 1 = 0 \) → \( A = \frac{1}{2} \)

### Step 7: Find \( \theta \)
Since \( A = \cos\theta \), we have:
\[ \cos\theta = \frac{1}{2} \]
The angle \( \theta \) that satisfies this in the interval \( 0^\circ < \theta < 90^\circ \) is:
\[ \theta = 60^\circ \]

### Final Answer:
\[ \theta = 60^\circ \]