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

Explanation

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^2 \theta - \cos \theta = \frac{1}{4} \]
Using the Pythagorean identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can substitute for \( \sin^2 \theta \):
\[ 1 - \cos^2 \theta - \cos \theta = \frac{1}{4} \]

### Step 2: Rearrange the equation
Next, we rearrange the equation to isolate terms:
\[ 1 - \cos^2 \theta - \cos \theta - \frac{1}{4} = 0 \]
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: Use the quadratic formula
This is a quadratic equation in terms of \( \cos \theta \). We can let \( x = \cos \theta \) and rewrite it as:
\[ x^2 + x - \frac{3}{4} = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -\frac{3}{4} \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -\frac{3}{4}}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 3}}{2} \]
\[ x = \frac{-1 \pm \sqrt{4}}{2} \]
\[ x = \frac{-1 \pm 2}{2} \]

### Step 4: Solve for \( x \)
Calculating the two possible values:
1. \( x = \frac{1}{2} \)
2. \( x = -\frac{3}{2} \) (This value is not valid since \( \cos \theta \) must be in the range \([-1, 1]\))

Thus, we have:
\[ \cos \theta = \frac{1}{2} \]

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

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