Find all real numbers that satisfy the equation sin x = -sqrt(2)/2.

Step-by-step solution

1. 1

   Find the reference angle from the magnitude. Ignore the sign for a moment: $\sin\theta = \tfrac{\sqrt2}{2}$ at $\theta = \tfrac{\pi}{4}$ (the $45^{\circ}$ angle of the isosceles right triangle). So the reference angle is $\tfrac{\pi}{4}$.

2. 2

   Use the sign to choose quadrants. Sine is the $y$-coordinate on the unit circle, so $\sin x < 0$ puts $x$ in quadrant III or IV. There are exactly two such angles per revolution, and both must appear in the answer.

3. 3

   Write the two base angles. In quadrant III the angle is $\pi + \tfrac{\pi}{4}$; in quadrant IV it is $2\pi - \tfrac{\pi}{4}$:

   $x = \frac{5\pi}{4} \qquad \text{and} \qquad x = \frac{7\pi}{4}$

4. 4

   Extend by the period. Sine has period $2\pi$, so add $2\pi n$ to each base angle:

   $\left\{x \;\middle|\; x = \frac{5\pi}{4} + 2\pi n \ \text{ or } \ x = \frac{7\pi}{4} + 2\pi n,\ n \in \mathbb{Z}\right\}$

5. 5

   Verify both families. $\sin\tfrac{5\pi}{4} = -\tfrac{\sqrt2}{2}$ ✓ and $\sin\tfrac{7\pi}{4} = -\tfrac{\sqrt2}{2}$ ✓. Distractor sets built on $\tfrac{4\pi}{3}$ and $\tfrac{5\pi}{3}$ fail, because those give $\sin x = -\tfrac{\sqrt3}{2} \approx -0.866$, not $-0.7071$. And a set built on $\tfrac{3\pi}{4}$ fails because that angle is in quadrant II, where sine is positive.