Find all real numbers that satisfy sin x = √3/2.

Step-by-step solution

1. 1

   Find the reference angle. The reference angle is the acute angle whose sine has the same magnitude:

   $\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\quad\Longrightarrow\quad \text{reference angle}=\frac{\pi}{3}$

2. 2

   Decide which quadrants apply. The value $\tfrac{\sqrt3}{2}$ is positive, and $\sin x>0$ in quadrants I and II. So there are exactly two solutions in one period $[0,2\pi)$ — a sine equation with a value strictly between $-1$ and $1$ always has two.

3. 3

   Write the two base solutions. Quadrant I gives the reference angle itself; quadrant II gives $\pi$ minus it:

   $x=\frac{\pi}{3},\qquad x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

4. 4

   Extend by the period. Sine has period $2\pi$, so every solution is obtained by adding integer multiples of $2\pi$:

   $\boxed{\left\{x\ \middle|\ x=\frac{\pi}{3}+2\pi n\ \text{ or }\ x=\frac{2\pi}{3}+2\pi n,\ n\in\mathbb{Z}\right\}}$

5. 5

   Check both families and note the compact form. $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ ✓ and $\sin\tfrac{2\pi}{3}=\sin\left(\pi-\tfrac{\pi}{3}\right)=\tfrac{\sqrt3}{2}$ ✓. The two families can also be written as the single expression $x=(-1)^{n}\tfrac{\pi}{3}+\pi n$, $n\in\mathbb{Z}$, which generates the same set. Adding $\pi n$ instead of $2\pi n$ to $\tfrac{\pi}{3}$ alone would wrongly include $x=\tfrac{4\pi}{3}$, where $\sin x=-\tfrac{\sqrt3}{2}$.