Solve for θ in: sin(2θ - 10°) = -0.5 where 0° ≤ θ ≤ 360°

Step 1: Start with the given equation.
$0.5 \tan^2 B - 0.5 = 0$

Step 2: Add $0.5$ to both sides.
$0.5 \tan^2 B = 0.5$

Step 3: Divide both sides by $0.5$.
$\tan^2 B = 1$

Step 4: Take the square root of both sides.
$\tan B = \pm 1$

Step 5: Solve $\tan B = 1$ for $B \in [0^\circ, 360^\circ)$.
The reference angle is $45^\circ$.
General solution: $B = 45^\circ + 180^\circ k$, $k \in \mathbb{Z}$.
$B = 45^\circ, \quad 45^\circ + 180^\circ = 225^\circ$

Step 6: Solve $\tan B = -1$ for $B \in [0^\circ, 360^\circ)$.
The reference angle is $45^\circ$.
Solutions in quadrants II and IV:
$B = 180^\circ - 45^\circ = 135^\circ, \quad 360^\circ - 45^\circ = 315^\circ$

Step 7: All solutions:
$B = 45^\circ, 135^\circ, 225^\circ, 315^\circ$

45^\circ, 135^\circ, 225^\circ, 315^\circ