Euler's Eyes
2026-10-20
Equations: looking for hidden angles
In the interactive book: complete chapter VII, with all demonstrations and exercises.
When does the cabin reach the same height?
Looking for $\operatorname{sen} x=1/2$ is asking in which positions the cabin is at height $1/2$. A horizontal line cuts the cycle at two points: one in the first quadrant and one in the second. In one turn, these are $30°$ and $150°$.
In this chapter, the main examples use degrees. When $\pi$ appears, the angle will be in radians.
Demo: two response families
The point of $\alpha$ and $180°-\alpha$ have the same height. Also, full turns keep the point. These are all possibilities because a horizontal line cuts a circle at a maximum of two points. Logo:
$$\operatorname{sen} x=\operatorname{sen}\alpha\iff \begin{cases} x=\alpha+360°k,\ \text{ou}\\ x=180°-\alpha+360°k. \end{cases}$$For the cosine, a vertical line gives the points $\alpha$ and $-\alpha$:
$$\cos x=\cos\alpha\iff x=\pm\alpha+360°k.$$For tangent, opposite directions have the same coordinate quotient:
$$\tan x=\tan\alpha\iff x=\alpha+180°k,$$with defined tangents. Here $k$ is an integer. To use radians, change $360°$ to $2\pi$ and $180°$ to $\pi$.
Example with range
Solve $\operatorname{sen} x=1/2$ into $0°\le x<720°$. From the first family come $30°$ and $390°$; of the second, $150°$ and $510°$. The break allows for two laps. A complete answer respects the ends of the range.
If the requested value for sine or cosine is outside $[-1,1]$, there is no real solution. For example, $\cos x=2$ is impossible.
Equations that change clothes
Consider $2\operatorname{sen}^2x-\operatorname{sen} x=0$. Highlight $\operatorname{sen} x$:
$$\operatorname{sen} x(2\operatorname{sen} x-1)=0.$$A product is zero when at least one factor is zero. So, $\operatorname{sen} x=0$ or $\operatorname{sen} x=1/2$. In $[0°,360°)$, the solutions are $0°$, $30°$, $150°$, and $180°$.
If we divided the equation by $\operatorname{sen} x$ from the beginning, we would lose the zero sine solutions. Factoring keeps possibilities visible. After squaring an equation, check each answer in the initial equation, as squaring can add solutions.
Inverse Functions: Choose a Top Answer
Sine does not have an inverse on the entire line, as multiple inputs produce the same output. To define arc sine, we restrict the angle to $[-90°,90°]$. In this range, each value of $[-1,1]$ appears only once.
| Inverse function | Prohibited | Main output |
|---|---|---|
| $\arcsin y$ | $-1\le y\le1$ | $[-90°,90°]$ |
| $\arccos y$ | $-1\le y\le1$ | $[0°,180°]$ |
| $\arctan y$ | any real | $(-90°,90°)$ |
Thus, $\arcsin(1/2)=30°$, but the equation $\operatorname{sen} x=1/2$ has infinitely many solutions. The inverse provides the main value, which is used to build families. $\arcsin$ also does not mean $1/\operatorname{sen}$: this multiplicative inverse is the cosecant.
To obtain $\arcsin(\operatorname{sen}150°)$, we first calculate $\operatorname{sen}150°=1/2$. The main output is $30°$, not $150°$.
In everyday life
- When does the cabin get there? Asking when the cabin of a Ferris wheel reaches a certain height is solving a sine equation.
- Tides. The times of high tide and low tide come from trigonometric equations, and are repeated with a period of around 12 h 25 min.
- Daylight. The days of the year with exactly 12 hours of light (the equinoxes) are the solutions of a periodic equation.
Exercises
Try solving it in your notebook before opening the solution. Exercises 1 and 2 are input; 3 and 4, for practice; 5 and 6, for deepening.
1. Solve $\cos x=1/2$ into $[0°,360°)$.
See resolution 1
$60°$ and $300°$, as they are the two horizontal coordinate points $1/2$.
2. Solve $\tan x=-1$ into $[0°,360°)$.
See resolution 2
$135°$ and $315°$, which differ by $180°$.
3. Give all real solutions of $\operatorname{sen} x=0$ in radians.
See resolution 3
$x=k\pi$, with $k$ integer. They are the points at the two extremes of the horizontal axis.
4. Solve $\operatorname{sen}(2x)=0$ into $[0°,360°)$.
See resolution 4
$2x=180°k$, then $x=90°k$. In the range: $0°$, $90°$, $180°$ and $270°$.
5. Calculate $\arccos(-1/2)$ in degrees.
See resolution 5
$120°$, which is in the main track $[0°,180°]$.
6. Solve $2\cos^2x-1=0$ into $[0°,360°)$.
See resolution 6
$\cos^2x=1/2$, then $\cos x=\pm\sqrt2/2$. Solutions: $45°$, $135°$, $225°$ and $315°$.