Solve the following equation:

$$
\sin x + \cos x = \sqrt{\frac{3}{2}}.
$$
I. Solution: Square both sides:

$$
\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=\frac{3}{2}
$$

which is

$$
1+\sin 2 x=\frac{3}{2}
$$

from which

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

and thus

$$
2 x_{1}=30^{\circ} \pm k \cdot 360^{\circ} \quad \text { and } \quad 2 x_{2}=150^{\circ} \pm k \cdot 360^{\circ}
$$

where $k=0,1,2, \ldots$

Therefore

$$
x_{1}=15^{\circ} \pm k \cdot 180^{\circ} \quad \text { and } \quad x_{2}=75^{\circ} \pm k \cdot 180^{\circ}
$$

These roots satisfy the equation (2), but it is not certain that they also satisfy the equation (1), because squaring can introduce extraneous solutions. It is only true that every root of (1) is necessarily a root of (2).

By substitution, we can verify that $x_{1}=195^{\circ}$ and $x_{2}=255^{\circ}$ do not satisfy the original equation (1), because both terms on the left side are negative, while the right side is positive. (By convention, every square root without a sign in front is considered positive. If we want to preserve the ambiguity of the square root, we must write $\pm$ in front of the square root, as we do in the quadratic formula.)

Therefore, the solution to our equation is

$$
x_{1}=15^{\circ} \pm k \cdot 360^{\circ}=\frac{\pi}{12} \pm 2 k \pi
$$

and

$$
x_{2}=75^{\circ} \pm k \cdot 360^{\circ}=\frac{5 \pi}{12} \pm 2 k \pi
$$

where $k=0,1,2, \ldots$

Rácz Márton (Bp. II., Rákóczi g. III. o. t.)

II. Solution: Since $\sin \left(x+45^{\circ}\right)=\sin x \cos 45^{\circ}+\cos x \sin 45^{\circ}=\frac{1}{\sqrt{2}}(\sin x+\cos x)$, we have $\sin x+\cos x=\sqrt{2} \sin \left(x+45^{\circ}\right)$

Thus, our equation can also be written as:

$$
\begin{gathered}
\sqrt{2} \sin \left(x+45^{\circ}\right)=\frac{\sqrt{3}}{\sqrt{2}}, \text { or } \sin \left(x+45^{\circ}\right)=\frac{\sqrt{3}}{2}, \text { from which } \\
x_{1}+45^{\circ}=60^{\circ} \pm k \cdot 360^{\circ}, \quad \text { and } \quad x_{2}+45^{\circ}=120^{\circ} \pm k \cdot 360^{\circ}, \quad(k=0,1,2, \ldots)
\end{gathered}
$$

and thus

$$
x_{1}=15^{\circ} \pm k \cdot 360^{\circ}, \quad x_{2}=75^{\circ} \pm k \cdot 360^{\circ} \quad \text { where } \quad k=0,1,2, \ldots
$$

Sóti Ferenc (Szeged, 5. sz. vegyip. techn. IV. o. t.)

III. Solution: Since $\cos \alpha=\sin \left(\alpha+\frac{\pi}{2}\right)$, our equation can also be written as:

$$
\sin x+\sin \left(x+\frac{\pi}{2}\right)=\sqrt{\frac{3}{2}}
$$

Using the identity $\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$

$$
2 \sin \left(x+\frac{\pi}{4}\right) \cos \frac{\pi}{4}=\frac{\sqrt{3}}{\sqrt{2}}
$$

But $\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}$ and thus

$$
\sin \left(x+\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}
$$

and so on, as in the second solution.