Intersection of a trigonometric function and a line

Finding where a trigonometric function meets a line amounts to solving a trigonometric equation. We find where $y = \sin x$ meets the line $y = \dfrac{1}{2}$.

Finding the intersections

At an intersection $\sin x = \dfrac{1}{2}$. On $0 \leq x < 2\pi$ two values satisfy it.

• $x = \dfrac{\pi}{6}$
• $x = \dfrac{5\pi}{6}$

The sine takes the value $\dfrac{1}{2}$ at $x = \dfrac{\pi}{6}$ and at the mirror position $x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.

Why two solutions appear

On the unit circle, the point at angle $\theta$ and the point at angle $\pi - \theta$ are mirror images across the $y$-axis. Since the sine is the height of the point, two points symmetric left and right have the same height.

That symmetry is why a single value produces two solutions.

Repetition by the period

Since $\sin x$ repeats with period $2\pi$, these two are not the only intersections. The same crossings recur every $2\pi$, giving infinitely many solutions.

Here $n$ is any integer.

The inverse function and the principal value

The inverse function gives $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$, but that is the principal value, taken from the restriction $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$, and it is only one of the answers. The rest are assembled from the symmetry and the periodicity just described.

For $\cos x = \dfrac{1}{2}$ the axis of symmetry moves from the $y$-axis to the $x$-axis, and the solutions take the form $x = \pm\dfrac{\pi}{3} + 2n\pi$.

The height of the line and the number of intersections

Questions such as finding the stretches of time during which an alternating voltage exceeds a given level reduce to equations of exactly this kind.

The large dots are the two intersections on $0 \leq x < 2\pi$, namely $\left( \dfrac{\pi}{6}, \dfrac{1}{2} \right)$ and $\left( \dfrac{5\pi}{6}, \dfrac{1}{2} \right)$.