[Algebra]
Problem: Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
If $k=-0.7$, determine all solutions to the equation $f(x)=0$.

Solution: Since $\sin ^{2} x+\cos ^{2} x=1$, then $\cos ^{2} x=1-\sin ^{2} x$, so

$$
\begin{aligned}
f(x) & =\sin ^{6} x+\left(1-\sin ^{2} x\right)^{3}+k\left(\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}\right) \\
& =\sin ^{6} x+1-3 \sin ^{2} x+3 \sin ^{4} x-\sin ^{6} x+k\left(\sin ^{4} x+1-2 \sin ^{2} x+\sin ^{4} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x+(3+2 k) \sin ^{4} x
\end{aligned}
$$

Now, we have

$$
f(x)=(1+k)-(3+2 k) \sin ^{2} x+(3+2 k) \sin ^{4} x
$$

and so we want to solve

$$
\begin{array}{r}
0.3-(1.6) \sin ^{2} x+(1.6) \sin ^{4} x=0 \\
16 \sin ^{4} x-16 \sin ^{2} x+3=0 \\
\left(4 \sin ^{2} x-3\right)\left(4 \sin ^{2} x-1\right)=0
\end{array}
$$

Therefore, $\sin ^{2} x=\frac{1}{4}, \frac{3}{4}$, and so $\sin x= \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}$.

Therefore,

$$
x=\frac{1}{6} \pi+2 \pi k, \frac{5}{6} \pi+2 \pi k, \frac{7}{6} \pi+2 \pi k, \frac{11}{6} \pi+2 \pi k, \frac{1}{3} \pi+2 \pi k, \frac{2}{3} \pi+2 \pi k, \frac{4}{3} \pi+2 \pi k, \frac{5}{3} \pi+2 \pi k
$$

for $k \in \mathbb{Z}$.
Since $\sin ^{2} x+\cos ^{2} x=1$, then

$$
\begin{aligned}
f(x) & =\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)+k\left(\sin ^{4} x+\cos ^{4} x\right) \\
& =\left(\sin ^{4}+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-3 \sin ^{2} x \cos ^{2} x\right) \\
& \quad k\left(\sin ^{4} x+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-2 \sin ^{2} x \cos ^{2} x\right) \\
& =\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x\right)+k\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right) \\
& =1-3 \sin ^{2} x \cos ^{2} x+k\left(1-2 \sin ^{2} x \cos ^{2} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x \cos ^{2} x
\end{aligned}
$$

Now, we have

$$
f(x)=(1+k)-(3+2 k) \sin ^{2} x \cos ^{2} x
$$

Using the fact that $\sin 2 x=2 \sin x \cos x$, we can further simplify $f(x)$ to

$$
f(x)=(1+k)-\frac{1}{4}(3+2 k) \sin ^{2} 2 x
$$



and so we want to solve

$$
\begin{aligned}
0.3-\frac{1}{4}(1.6) \sin ^{2} 2 x & =0 \\
4 \sin ^{2} 2 x & =3 \\
\sin ^{2} 2 x & =\frac{3}{4}
\end{aligned}
$$

and so $\sin 2 x= \pm \frac{\sqrt{3}}{2}$.

Therefore,

$$
2 x=\frac{1}{3} \pi+2 \pi k, \frac{2}{3} \pi+2 \pi k, \frac{4}{3} \pi+2 \pi k, \frac{5}{3} \pi+2 \pi k
$$

for $k \in \mathbb{Z}$, and so

$$
x=\frac{1}{6} \pi+\pi k, \frac{1}{3} \pi+\pi k, \frac{2}{3} \pi+\pi k, \frac{5}{6} \pi+\pi k
$$

for $k \in \mathbb{Z}$.

Final answer: $x=\frac{1}{6} \pi+\pi k, \frac{1}{3} \pi+\pi k, \frac{2}{3} \pi+\pi k, \frac{5}{6} \pi+\pi k$