Let's find the numbers $x$ between 0 and 30 for which the sine of $x$ degrees is equal to the sine of $x$ radians. - How many such numbers are there between 30 and 90?
I. The radian measure of an angle of $x$ degrees is $\pi x / 180$, hence the requirement

$$
\sin x=\sin \frac{\pi x}{180}
$$

The sines of two angles (rotation angles) are equal if and only if their difference is an integer multiple of $2 \pi$, or if their sum is an odd multiple of $\pi$, that is,

$$
x-\frac{\pi x}{180}=2 k \pi, \quad \text { or } \quad x+\frac{\pi x}{180}=(2 k+1) \pi
$$

where clearly $k \geq 0$. From this,

$$
x=\frac{360 \pi}{180-\pi} \cdot k \quad \text { or } \quad x=\frac{180 \pi}{180+\pi} \cdot(2 k+1)
$$

If we use the known approximation $22 / 7$ for $\pi$ (with a $0.04\%$ error),

$$
\begin{aligned}
& x \approx \frac{3960}{619} k \approx 6.397 k, \quad \text { or } \\
& x \approx \frac{1980}{641}(2 k+1)=\frac{1980}{641}+\frac{3960}{641} k \approx 3.088 \cdot(2 k+1)
\end{aligned}
$$

In the first case, with $0 \leq k \leq 4$,

$$
x \approx 0 ; \quad 6.397 ; \quad 12.79 ; \quad 19.19 ; \quad 25.59
$$

In the second case, with the same values,

$$
x \approx 3.088 ; \quad 9.264 ; \quad 15.44 ; \quad 21.62 ; \quad 27.79
$$

The number of corresponding values is 10.

II. Due to the constraint $x<90$,

(1) gives: $\quad k<\frac{619}{44}<14.1, \quad$ that is, $\quad k \leq 14$,

(2) gives: $\quad 2 k+1<\frac{641}{22}<29.2, \quad$ that is, $\quad k \leq 14$,

so both formulas give $14-4=10$ valid $x$ values.

János Pintér (Budapest, I. István g. III. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_6f623c6a2ffceb3bfa69g-1.jpg?height=655&width=1379&top_left_y=1624&top_left_x=362)

Note. The corresponding $x$ values can be read off as the abscissas of the intersection points of the curves representing the functions

$$
y=\sin x \quad \text { and } \quad y=\sin \frac{\pi}{180} \cdot x
$$

in the interval $0 \leq x \leq 30$. (On the graph, the unit on the ordinate axis is 10 times larger than that on the abscissa axis.)