Solve `sin^(2) theta-cos theta=1/4, 0 le theta le 2pi`.

Explanation

To solve the equation \( \sin^2 \theta - \cos \theta = \frac{1}{4} \) for \( 0 \leq \theta \leq 2\pi \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin^2 \theta - \cos \theta = \frac{1}{4} \]
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can rewrite the equation in terms of cosine:
\[ 1 - \cos^2 \theta - \cos \theta = \frac{1}{4} \]

### Step 2: Rearrange the equation
Next, we rearrange the equation:
\[ 1 - \cos^2 \theta - \cos \theta - \frac{1}{4} = 0 \]
This simplifies to:
\[ -\cos^2 \theta - \cos \theta + \frac{3}{4} = 0 \]
Multiplying through by -1 gives:
\[ \cos^2 \theta + \cos \theta - \frac{3}{4} = 0 \]

### Step 3: Solve the quadratic equation
This is a quadratic equation in terms of \( \cos \theta \). We can use the quadratic formula \( \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 1 \), and \( c = -\frac{3}{4} \).
Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot \left(-\frac{3}{4}\right) = 1 + 3 = 4 \]
Now applying the quadratic formula:
\[ \cos \theta = \frac{-1 \pm \sqrt{4}}{2 \cdot 1} = \frac{-1 \pm 2}{2} \]
This gives us two possible solutions:
\[ \cos \theta = \frac{1}{2} \quad \text{and} \quad \cos \theta = -\frac{3}{2} \]

### Step 4: Evaluate the solutions
Since \( \cos \theta = -\frac{3}{2} \) is not possible (the range of cosine is \([-1, 1]\)), we only consider:
\[ \cos \theta = \frac{1}{2} \]

### Step 5: Find the angles
The angles for which \( \cos \theta = \frac{1}{2} \) in the interval \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{5\pi}{3} \]

### Conclusion
Thus, the solutions to the equation \( \sin^2 \theta - \cos \theta = \frac{1}{4} \) in the interval \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{3}, \quad \frac{5\pi}{3} \]