The equation ` sin^(4) x + cos^(4) x + sin 2x + k = 0 ` must have real solutions if :

Options

A. ` k = 0 `
B. ` |k| le (1)/(2) `
C. ` - (3)/(2) le k le (1)/(2) `
D. ` - (1)/(2) le k le (3)/(2) `

Answer

To solve the equation \( \sin^4 x + \cos^4 x + \sin 2x + k = 0 \) and determine the conditions under which it has real solutions, we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \]
Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify:
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]
Also, we can express \( \sin 2x \) as:
\[ \sin 2x = 2\sin x \cos x \]
Thus, we can rewrite the original equation as:
\[ 1 - 2\sin^2 x \cos^2 x + 2\sin x \cos x + k = 0 \]

### Step 2: Substitute \( \sin 2x \)
Let \( y = \sin 2x = 2\sin x \cos x \). Then \( \sin^2 x \cos^2 x = \frac{y^2}{4} \). The equation becomes:
\[ 1 - \frac{y^2}{2} + y + k = 0 \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ -\frac{y^2}{2} + y + (k + 1) = 0 \]
Multiplying through by -2 to eliminate the fraction:
\[ y^2 - 2y - 2(k + 1) = 0 \]

### Step 4: Determine the discriminant
For the quadratic equation \( ay^2 + by + c = 0 \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = -2 \), and \( c = -2(k + 1) \):
\[ D = (-2)^2 - 4(1)(-2(k + 1)) = 4 + 8(k + 1) \]
Setting the discriminant \( D \geq 0 \) for real solutions:
\[ 4 + 8(k + 1) \geq 0 \]

### Step 5: Solve the inequality
\[ 8(k + 1) \geq -4 \]
\[ k + 1 \geq -\frac{1}{2} \]
\[ k \geq -\frac{3}{2} \]

### Step 6: Conclusion
Thus, the equation \( \sin^4 x + \cos^4 x + \sin 2x + k = 0 \) must have real solutions if:
\[ k \geq -\frac{3}{2} \]