Which angles satisfy the following equation.

$$
\operatorname{cot}^{2} x-\sin ^{2} x=\frac{1}{4}
$$
$$
\operatorname{cotg}^{2} x=\frac{\cos ^{2} x}{\sin ^{2} x}=\frac{1-\sin ^{2} x}{\sin ^{2} x}
$$

Substituting this expression for $\operatorname{cotg}^{2} x$ into the equation, we get:

$$
\frac{1-\sin ^{2} x}{\sin ^{2} x}-\sin ^{2} x=\frac{1}{4} \quad \text { or } \quad 4 \sin ^{4} x+5 \sin ^{2} x-4=0
$$

This equation is quadratic in $\sin ^{2} x$, and it has two roots of opposite signs; of these, only the positive one can be valid, that is

$$
\sin ^{2} x=\frac{-5+\sqrt{89}}{8} \quad \text { and } \quad \sin x= \pm \sqrt{\frac{-5+\sqrt{89}}{8}}
$$

The angles corresponding to the positive value of $\sin x$, between $0^{\circ}$ and $360^{\circ}$:

$$
48^{\circ} 6^{\prime} 52^{\prime \prime} \quad \text { and } \quad 180^{\circ}-48^{\circ} 6^{\prime} 52^{\prime \prime}
$$

The angles corresponding to the negative value of $\sin x$, between $0^{\circ}$ and $360^{\circ}$:

$$
180^{\circ}+48^{\circ} 6^{\prime} 52^{\prime \prime} \text { and } 360^{\circ}-48^{\circ} 6^{\prime} 52^{\prime \prime}
$$

These - and all - solutions can be summarized in the following two sets of values:

$$
48^{\circ} 6^{\prime} 52^{\prime \prime}+k \cdot 180^{\circ} \quad \text { and } \quad-48^{\circ} 6^{\prime} 52^{\prime \prime}+k \cdot 180^{\circ}
$$

where $k$ represents any integer.

Tassonyi Kenéz (Áll. Toldy Ferenc r. VI. o. Bp. II.)