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`

Explanation

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 knowing that \( \sin^2 x + \cos^2 x = 1 \), we can rewrite it as:
...