If `sin^(2)x-cosx=1//4`, then the values of x between 0 and `2pi` are

Options

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

Answer

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

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

### Step 2: Rearrange the equation
Now, 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 the entire equation by -1 gives:
\[ \cos^2 x + \cos x - \frac{3}{4} = 0 \]

### Step 3: Multiply through by 4
To eliminate the fraction, we can multiply the entire equation by 4:
\[ 4\cos^2 x + 4\cos x - 3 = 0 \]

### Step 4: Factor the quadratic equation
Next, we will factor the quadratic equation:
\[ (4\cos x - 3)(\cos x + 1) = 0 \]

### Step 5: Solve for \( \cos x \)
Setting each factor to zero gives us:
1. \( 4\cos x - 3 = 0 \)
\[ \cos x = \frac{3}{4} \]
2. \( \cos x + 1 = 0 \)
\[ \cos x = -1 \]

### Step 6: Find the angles for \( \cos x = \frac{3}{4} \)
To find the values of \( x \) for \( \cos x = \frac{3}{4} \), we can use the inverse cosine function:
\[ x = \cos^{-1}\left(\frac{3}{4}\right) \]
This will give us one angle in the range \( [0, 2\pi] \). The second angle can be found using:
\[ x = 2\pi - \cos^{-1}\left(\frac{3}{4}\right) \]

### Step 7: Find the angles for \( \cos x = -1 \)
The angle for \( \cos x = -1 \) is:
\[ x = \pi \]

### Step 8: Compile all solutions
Thus, the values of \( x \) in the interval \( [0, 2\pi] \) are:
\[ x = \cos^{-1}\left(\frac{3}{4}\right), \quad x = 2\pi - \cos^{-1}\left(\frac{3}{4}\right), \quad x = \pi \]

### Final Answer
The values of \( x \) between \( 0 \) and \( 2\pi \) are:
\[ x = \cos^{-1}\left(\frac{3}{4}\right), \quad x = 2\pi - \cos^{-1}\left(\frac{3}{4}\right), \quad x = \pi \]

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