Solve sin(x) = -1/2 for all real x.

Step-by-step solution

1. 1

   Find the reference angle from the magnitude. Ignore the sign for a moment: $\sin\theta=\tfrac12$ at the special angle

   $\theta=\frac{\pi}{6}=30^\circ$

   from the $30$-$60$-$90$ triangle. So $\tfrac{\pi}{6}$ is the reference angle — no calculator needed.

2. 2

   Determine which quadrants have negative sine. Sine is the $y$-coordinate on the unit circle, so it is negative in Quadrant III and Quadrant IV (below the axis). Two quadrants means two solutions per revolution.

3. 3

   Build the Quadrant III solution. Measure the reference angle past $\pi$:

   $x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}$

   Here $\sin\tfrac{7\pi}{6}=-\tfrac12$ ✓ (both coordinates negative).

4. 4

   Build the Quadrant IV solution. Measure the reference angle back from $2\pi$:

   $x=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}$

   Here $\sin\tfrac{11\pi}{6}=-\tfrac12$ ✓.

5. 5

   Extend by the period. Sine has period $2\pi$, so every solution is one of these two shifted by whole revolutions:

   $x=\frac{7\pi}{6}+2\pi k\qquad\text{or}\qquad x=\frac{11\pi}{6}+2\pi k,\qquad k\in\mathbb{Z}$

   Both families are needed: they are $\tfrac{4\pi}{6}=\tfrac{2\pi}{3}$ apart, not a multiple of $2\pi$. (Equivalently, $x=-\tfrac{\pi}{6}+2\pi k$ describes the same Quadrant IV family.)

6. 6

   Verify across several k. Evaluating both families at $k=-2,\,0,\,3$ gives $\sin x=-0.500000$ in all six cases ✓, confirming the period and both branches.