Solve the following equation:

$$
\sin 2 x=\cos 3 x .
$$
Since $\sin 2 x=2 \sin x \cos x$, and $\cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\left(1-2 \sin ^{2} x\right) \cos x-2 \sin ^{2} x \cos x=$ $\cos x\left(1-4 \sin ^{2} x\right)$, our equation, reduced to 0, can be written as:

$$
2 \sin x \cos x-\cos x\left(1-4 \sin ^{2} x\right)=0
$$

or

$$
\cos x\left(4 \sin ^{2} x+2 \sin x-1\right)=0
$$

Here, either

$$
\cos x=0
$$

or

$$
4 \sin ^{2} x+2 \sin x-1=0
$$

In the first case,

$$
x_{1}=90^{\circ} \pm k \cdot 360^{\circ}, \quad x_{2}=270^{\circ} \pm k \cdot 360^{\circ} \quad(k=0,1,2, \ldots)
$$

In the second case,

$$
\sin x=\frac{-2 \pm \sqrt{4+16}}{8}=\frac{-1 \pm \sqrt{5}}{4}
$$

Calculating the approximate values, the table gives $x$ as $18^{\circ}$, or $-54^{\circ}$, but it is certainly questionable whether these values are exact.

We can resolve this issue by calculating the values of $\sin 18^{\circ}$ and $\sin 54^{\circ}$. This can be done, for example, as follows:

Using known trigonometric identities,

$$
\begin{gathered}
2 \sin 36^{\circ} \cos 36^{\circ}=\sin 72^{\circ} \\
2 \sin 72^{\circ} \cos 72^{\circ}=\sin 144^{\circ}=\sin 36^{\circ}
\end{gathered}
$$

Multiplying these two identities and simplifying by $\sin 36^{\circ} \sin 72^{\circ}$,

$$
4 \cos 36^{\circ} \cos 72^{\circ}=4 \sin 54^{\circ} \sin 18^{\circ}=1
$$

Considering (1),

$$
\sin 54^{\circ}-\sin 18^{\circ}=2 \cos 36^{\circ} \sin 18^{\circ}=2 \sin 54^{\circ} \sin 18^{\circ}=\frac{1}{2}
$$

Furthermore,

$$
\left(\sin 54^{\circ}+\sin 18^{\circ}\right)^{2}=\left(\sin 54^{\circ}-\sin 18^{\circ}\right)^{2}+4 \sin 54^{\circ} \sin 18^{\circ}
$$

and thus, using (2) and (1),

$$
\sin 54^{\circ}+\sin 18^{\circ}=\sqrt{\frac{1}{4}+1}=\frac{\sqrt{5}}{2}
$$

From (2) and (3),

$$
\sin 18^{\circ}=\frac{-1+\sqrt{5}}{4} \quad \text { and } \quad \sin 54^{\circ}=\frac{1+\sqrt{5}}{4}
$$

Therefore,

$$
\begin{array}{cc}
x_{3}=18^{\circ} \pm k \cdot 360^{\circ}, & x_{4}=162^{\circ} \pm k \cdot 360^{\circ} \\
x_{5}=234^{\circ} \pm k \cdot 360^{\circ}, & x_{6}=306^{\circ} \pm k \cdot 360^{\circ} \\
\text { where } k=0,1,2, \ldots
\end{array}
$$

By substitution, we can verify that all six principal values satisfy our equation.

Papp Kálmán (Bp. IX., Fáy A. g. IV. o. t.)

Second solution: We can write our equation as:

$$
\sin 2 x=\sin \left(90^{\circ}-3 x\right)
$$

This equation holds in two cases, if

$$
2 x=90^{\circ}-3 x \pm k \cdot 360^{\circ} \quad \text { or } \quad 180^{\circ}-2 x=90^{\circ}-3 x \pm k \cdot 360^{\circ}
$$

From the first,

$$
5 x=90^{\circ} \pm k \cdot 360^{\circ}, \quad \text { i.e., } \quad x=18^{\circ} \pm k \cdot 72^{\circ}
$$

From this, depending on whether $k=5 l, 5 l+1,5 l+2,5 l+3,5 l+4$,

$$
\begin{aligned}
& x_{1}=18^{\circ} \pm l \cdot 360^{\circ}, \quad x_{2}=90^{\circ} \pm l \cdot 360^{\circ}, \quad x_{3}=162^{\circ} \pm l \cdot 360^{\circ} \\
& x_{4}=234^{\circ} \pm l \cdot 360^{\circ}, \quad x_{5}=306^{\circ} \pm l \cdot 360^{\circ} . \quad(l=0,1,2, \ldots)
\end{aligned}
$$

From the second equality,

$$
x_{6}=-90^{\circ} \pm k \cdot 360=270^{\circ} \pm k^{\prime} \cdot 360^{\circ}\left(\text { where } k^{\prime}=k-1, k=1,2, \ldots\right)
$$