if `cos^4x+sin^2x-p=0` has real solutions then

Options

A. `p le 1`
B. `(3)/(4) le p lt 1`
C. `p ge (3)/(4)`
D. none of these

Explanation

To solve the equation \( \cos^4 x + \sin^2 x - p = 0 \) for the values of \( p \) that allow for real solutions, we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \sin^2 x = 1 - \cos^2 x \). Therefore, we can substitute \( \sin^2 x \) in the equation:

\[ \cos^4 x + (1 - \cos^2 x) - p = 0 \]

### Step 2: Simplify the equation
Now, simplify the equation:

\[ \cos^4 x - \cos^2 x + 1 - p = 0 \]

Rearranging gives us:

\[ \cos^4 x - \cos^2 x + (1 - p) = 0 \]

### Step 3: Let \( y = \cos^2 x \)
Let \( y = \cos^2 x \). Then, we can rewrite the equation as a quadratic in \( y \):

\[ y^2 - y + (1 - p) = 0 \]

### Step 4: Apply the discriminant condition
For the quadratic equation \( ay^2 + by + c = 0 \) to have real solutions, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

In our case, \( a = 1 \), \( b = -1 \), and \( c = 1 - p \). Thus, the discriminant is:

\[ D = (-1)^2 - 4 \cdot 1 \cdot (1 - p) \geq 0 \]

This simplifies to:

\[ 1 - 4(1 - p) \geq 0 \]

### Step 5: Solve the inequality
Now, simplify the inequality:

\[ 1 - 4 + 4p \geq 0 \]

\[ 4p - 3 \geq 0 \]

\[ 4p \geq 3 \]

\[ p \geq \frac{3}{4} \]

### Step 6: Determine the upper limit for \( p \)
Since \( y = \cos^2 x \) must be in the interval \([0, 1]\), we need to ensure that the roots of the quadratic fall within this range.

1. Evaluate the quadratic at \( y = 0 \):
\[ 0^2 - 0 + (1 - p) = 1 - p > 0 \implies p < 1 \]

2. Evaluate the quadratic at \( y = 1 \):
\[ 1^2 - 1 + (1 - p) = 1 - p > 0 \implies p < 1 \]

Thus, we find that \( p \) must satisfy:

\[ \frac{3}{4} \leq p < 1 \]

### Conclusion
The values of \( p \) for which the equation \( \cos^4 x + \sin^2 x - p = 0 \) has real solutions are:

\[ \frac{3}{4} \leq p < 1 \]