Solve `sin^2x-cosx =1/4`, `0 le x le 2pi`

Options

A. `(2pi)/3, pi/3`
B. `pi/3, (5pi)/3`
C. `-pi/3 , (2pi)/3`
D. `(2pi)/3, (5pi)/3`

Answer

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

### Step 1: Rewrite the equation using the Pythagorean identity
We know that \( \sin^2 x + \cos^2 x = 1 \). Therefore, we can express \( \sin^2 x \) as:
\[ \sin^2 x = 1 - \cos^2 x \]
Substituting this into the original equation gives:
\[ 1 - \cos^2 x - \cos x = \frac{1}{4} \]

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

### Step 3: Solve the quadratic equation
Now we can treat this as a quadratic equation in terms of \( \cos x \). Let \( y = \cos x \):
\[ y^2 + y - \frac{3}{4} = 0 \]
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = 1, c = -\frac{3}{4} \).
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -\frac{3}{4}}}{2 \cdot 1} \]
\[ y = \frac{-1 \pm \sqrt{1 + 3}}{2} \]
\[ y = \frac{-1 \pm 2}{2} \]
This gives us two possible solutions:
\[ y = \frac{1}{2} \quad \text{or} \quad y = -\frac{3}{2} \]

### Step 4: Analyze the solutions
Since \( y = \cos x \), we discard \( y = -\frac{3}{2} \) because the cosine function only takes values in the range \([-1, 1]\).

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

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

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

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