Solve sin 2x = √3/2 for x in the interval from 0 to 2π.

Step-by-step solution

1. 1

   Track what the interval does to the double angle. If $x$ runs over $[0, 2\pi]$ then $2x$ runs over $[0, 4\pi]$ — two full turns. That is why this equation has twice as many solutions as $\sin x = \tfrac{\sqrt3}{2}$ would, and forgetting it is the standard way to lose half the answers.

2. 2

   Solve for the double angle in general form. $\sin\theta = \tfrac{\sqrt3}{2}$ has the reference angle $\tfrac{\pi}{3}$, with solutions in the first and second quadrants:

   $2x = \frac{\pi}{3} + 2\pi k \qquad \text{or} \qquad 2x = \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$

3. 3

   Halve to get the general solution for x. Dividing every term by $2$ turns the period $2\pi$ into $\pi$:

   $x = \frac{\pi}{6} + \pi k \qquad \text{or} \qquad x = \frac{\pi}{3} + \pi k$

   The halved period $\pi$ is the algebraic reason two families become four solutions on a $2\pi$-long interval.

4. 4

   Select the values that land inside [0, 2π]. From the first family: $k=0$ gives $\tfrac{\pi}{6}$, $k=1$ gives $\tfrac{7\pi}{6}$, and $k=2$ gives $\tfrac{13\pi}{6} \approx 6.807 > 2\pi \approx 6.283$, which is out. From the second: $k=0$ gives $\tfrac{\pi}{3}$, $k=1$ gives $\tfrac{4\pi}{3}$, and $k=2$ gives $\tfrac{7\pi}{3}$, also out. Negative $k$ gives negative angles.

5. 5

   List and verify the four solutions.

   $x = \frac{\pi}{6},\quad \frac{\pi}{3},\quad \frac{7\pi}{6},\quad \frac{4\pi}{3}$

   Doubling each gives $2x = \tfrac{\pi}{3},\ \tfrac{2\pi}{3},\ \tfrac{7\pi}{3},\ \tfrac{8\pi}{3}$, and $\sin$ of each equals $\tfrac{\sqrt3}{2} \approx 0.8660254$ to fifteen decimal places. Note the pairs $\left(\tfrac{\pi}{6}, \tfrac{\pi}{3}\right)$ and $\left(\tfrac{7\pi}{6}, \tfrac{4\pi}{3}\right)$ are symmetric about $\tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$, the peaks of $\sin 2x$.