The equation `sin^4x+cos^4x+sin2x+alpha=0` is solvable for `-5/2lt=alphalt=1/2` (b) `-3lt=alpha<1` `-3/2lt=alphalt=1/2` (d) `-1lt=alphalt=1`

Options

A. `-(1)/(2) le alpha (1)/(2)`
B. `-3 le alpha le 1`
C. `-(3)/(2) le alpha le (1)/(2)`
D. `-1 le alpha le 1`

Answer

To solve the equation \( \sin^4 x + \cos^4 x + \sin 2x + \alpha = 0 \) for the values of \( \alpha \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \sin^4 x + \cos^4 x + \sin 2x + \alpha = 0 \]
Using the identity \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \), and since \( \sin^2 x + \cos^2 x = 1 \), we can rewrite it as:
\[ 1 - 2\sin^2 x \cos^2 x + \sin 2x + \alpha = 0 \]

### Step 2: Substitute \( \sin 2x \)
Recall that \( \sin 2x = 2\sin x \cos x \). Thus, we can substitute this into the equation:
\[ 1 - 2\sin^2 x \cos^2 x + 2\sin x \cos x + \alpha = 0 \]

### Step 3: Let \( y = \sin 2x \)
Let \( y = \sin 2x \). Then \( \sin^2 x \cos^2 x = \frac{y^2}{4} \). Substituting this into the equation gives:
\[ 1 - 2\left(\frac{y^2}{4}\right) + y + \alpha = 0 \]
This simplifies to:
\[ 1 - \frac{y^2}{2} + y + \alpha = 0 \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ -\frac{y^2}{2} + y + (1 + \alpha) = 0 \]
Multiplying through by -2 to eliminate the fraction:
\[ y^2 - 2y - 2(1 + \alpha) = 0 \]

### Step 5: Finding the Discriminant
For this quadratic equation in \( y \) to have real solutions, the discriminant must be non-negative:
\[ D = (-2)^2 - 4 \cdot 1 \cdot (-2(1 + \alpha)) \geq 0 \]
Calculating the discriminant:
\[ D = 4 + 8(1 + \alpha) = 4 + 8 + 8\alpha = 12 + 8\alpha \geq 0 \]
This simplifies to:
\[ 8\alpha \geq -12 \quad \Rightarrow \quad \alpha \geq -\frac{3}{2} \]

### Step 6: Finding the Upper Bound
The maximum value of \( y = \sin 2x \) is 1. Therefore, substituting \( y = 1 \) into the quadratic equation:
\[ 1^2 - 2(1) - 2(1 + \alpha) = 0 \]
This gives:
\[ 1 - 2 - 2 - 2\alpha = 0 \quad \Rightarrow \quad -3 - 2\alpha = 0 \quad \Rightarrow \quad 2\alpha = -3 \quad \Rightarrow \quad \alpha = -\frac{3}{2} \]

### Step 7: Conclusion
Combining the inequalities, we find:
\[ -\frac{3}{2} \leq \alpha \leq \frac{1}{2} \]

Thus, the equation is solvable for:
\[ \boxed{-\frac{3}{2} \leq \alpha \leq \frac{1}{2}} \]