Solve the following equation:

$$
\sin ^{4} x+\cos ^{4} x=2 \sin 2 x .
$$
Given the identity $\sin ^{2} x+\cos ^{2} x=1$, the given equation can also be written as:

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

or

$$
1-\frac{1}{2} \sin ^{2} x=2 \sin 2 x
$$

or

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

from which

$$
\sin 2 x=-2 \pm \sqrt{6}
$$

Since the absolute value of $\sin 2 x$ is less than 1, the radical can only be taken with a positive sign, and thus

$$
\sin 2 x=-2+\sqrt{6}=0.44949
$$

From this equation, considering only acute angles:

$$
x_{1}=13^{\circ} 21^{\prime} 19^{\prime \prime} \quad \text { and } \quad x_{2}=76^{\circ} 38^{\prime} 41^{\prime \prime} \text {. }
$$

(Neuhold Özséb, Eger.)

Number of solutions: 49.