Solve the following equation:

$$
\sin ^{6} x+\cos ^{6} x=\frac{7}{16}
$$
I. Solution. In the left-hand side of our equation, we factor out $\sin ^{2} x+\cos ^{2} x=1$

$$
\begin{aligned}
\sin ^{6} x+\cos ^{6} x & =\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)= \\
& =\sin ^{4} x-\sin ^{2} x\left(1-\sin ^{2} x\right)+\left(1-\sin ^{2} x\right)^{2}= \\
& =\sin ^{4} x-\sin ^{2} x+\sin ^{4} x+1-2 \sin ^{2} x+\sin ^{4} x=3 \sin ^{4} x-3 \sin ^{2} x+1=\frac{7}{16}
\end{aligned}
$$

thus

$$
3 \sin ^{4} x-3 \sin ^{2} x+\frac{9}{16}=0
$$

or

$$
\sin ^{4} x-\sin ^{2} x+\frac{3}{16}=0
$$

from which

$$
\sin ^{2} x=\frac{1 \pm \frac{1}{2}}{2}=\frac{2 \pm 1}{4}
$$

and thus

$$
\sin ^{2} x_{1,2}=\frac{3}{4}, \quad \sin ^{2} x_{3,4}=\frac{1}{4}
$$

or

$$
\sin x_{1,2}= \pm \frac{\sqrt{3}}{2}, \quad \sin x_{3,4}= \pm \frac{1}{2}
$$

Therefore, the sought roots are

$$
\begin{gathered}
x_{1}=60^{\circ} \pm k \pi, \quad x_{2}=120^{\circ} \pm k \pi, \quad x_{3}=30^{\circ} \pm k \pi, \quad x_{4}=150^{\circ} \pm k \pi \\
(k=0,1,2, \ldots)
\end{gathered}
$$

Thus, 8 solutions can be found between $0^{\circ}$ and $360^{\circ}$.

All the roots can be summarized in the following formula:

$$
x= \pm 30^{\circ} \pm k \cdot 90^{\circ}= \pm(3 k \pm 1) 30^{\circ} \quad(k=0,1,2, \ldots)
$$

Zoltán Szatmári (Bp. VIII., Piarist g. II. o. t.)

# II. Solution: 

$$
\begin{aligned}
\left(\sin ^{2} x+\cos ^{2} x\right)^{3} & =\sin ^{6} x+3 \sin ^{4} x \cos ^{2} x+3 \sin ^{2} x \cos ^{4} x+\cos ^{6} x= \\
& =\sin ^{6} x+\cos ^{6} x+3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)
\end{aligned}
$$

Performing the

$$
\begin{gathered}
\sin ^{2} x+\cos ^{2} x=1 \text { and } \sin ^{6} x+\cos ^{6} x=\frac{7}{16} \text { substitutions, } \\
\text { we get } 1=\frac{7}{16}+3 \sin ^{2} x \cos ^{2} x
\end{gathered}
$$

or, rearranging,

$$
3 \sin ^{2} x \cos ^{2} x=\frac{9}{16}
$$

Multiplying by $\frac{4}{3}$

$$
(2 \sin x \cos x)^{2}=\frac{3}{4}
$$

from which

$$
2 \sin x \cos x=\sin 2 x= \pm \frac{\sqrt{3}}{2}
$$

Thus,

$$
2 x= \pm \frac{\pi}{3} \pm k \pi, \quad(k=0,1,2, \ldots)
$$

and thus

$$
x= \pm \frac{\pi}{6} \pm k \frac{\pi}{2}= \pm(3 k \pm 1) \frac{\pi}{6} . \quad(k=0,1,2, \ldots)
$$