Solve 2sin(x) - 1 = 0 for all real x.

Step-by-step solution

1. 1

   Isolate the sine. Add $1$ and divide by $2$:

   $2\sin x=1\qquad\Longrightarrow\qquad \sin x=\frac{1}{2}$

   Since $\tfrac12$ lies in $[-1,1]$, solutions exist.

2. 2

   Locate the angle in the first quadrant. $\tfrac12$ is a special value: the $30^\circ$-$60^\circ$-$90^\circ$ triangle gives $\sin\tfrac{\pi}{6}=\tfrac12$, so $x=\tfrac{\pi}{6}$ is one solution — no calculator needed.

3. 3

   Find the second solution in the same revolution. Sine is positive in both the first and second quadrants, and $\sin(\pi-\theta)=\sin\theta$. So the reflection

   $x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$

   also works. Missing this branch is the single most common error on sine equations.

4. 4

   Extend by the period. $\sin$ repeats every $2\pi$, so every solution is one of the two base angles shifted by a whole number of revolutions:

   $x=\frac{\pi}{6}+2\pi k\qquad\text{or}\qquad x=\frac{5\pi}{6}+2\pi k,\qquad k\in\mathbb{Z}$

5. 5

   See why the two families do not merge. The gap from $\tfrac{\pi}{6}$ to $\tfrac{5\pi}{6}$ is $\tfrac{2\pi}{3}$, which is not a multiple of $2\pi$, so neither family contains the other — both are genuinely needed.

6. 6

   Verify by substitution. $2\sin\!\left(\tfrac{\pi}{6}\right)-1=2(0.5)-1=0$ ✓ and $2\sin\!\left(\tfrac{5\pi}{6}\right)-1=2(0.5)-1=0$ ✓. Testing $k=-2$ and $k=3$ on both families also returns $0$ to machine precision ✓.