Solve the equation

$$
\sin ^{4} x+\cos ^{4} x-2 \sin 2 x+\frac{3}{4} \sin ^{2} 2 x=0
$$
The identity $\sin ^{2} x+\cos ^{2} x=1$ squared on both sides gives:

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

Using the relationship $\sin 2 x=2 \sin x \cos x$,

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

Substituting this into the equation to be solved and multiplying by 4 immediately,

$$
\sin ^{2} 2 x-8 \sin 2 x+4=0
$$

The two roots of this quadratic equation in $\sin 2 x$ are: $4 \pm 2 \sqrt{3}$. Only the value less than 1, $4-2 \sqrt{3} \approx$ 0.5359, is valid. By back-substituting the approximate values,

$$
2 x_{1}=32.4^{\circ} \pm n \cdot 360^{\circ}, \quad 2 x_{2}=147.6^{\circ} \pm n \cdot 360^{\circ}
$$

Thus,

$$
x_{1}=16.2^{\circ} \pm n \cdot 180^{\circ}, \quad x_{2}=73.8^{\circ} \pm n \cdot 180^{\circ} \quad(n=0,1,2, \ldots)
$$

This solves the problem.

Tatai Péter (Bp. XIV., I. István g. III. o. t.)

Note: We can also arrive at the above equation by using the identities

$$
\sin ^{2} x=\frac{1-\cos 2 x}{2}, \quad \cos ^{2} x=\frac{1+\cos 2 x}{2}
$$

and the identity $\cos ^{2} 2 x=1-\sin ^{2} 2 x$ to transform our equation. Additionally, the transformation

$$
\begin{gathered}
\sin ^{4} x+\cos ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x= \\
=\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x=1-\frac{1}{2} \sin ^{2} 2 x
\end{gathered}
$$

also leads to the same result.