[Algebra]
Problem: Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
Determine all real numbers $k$ for which there exists a real number $c$ such that $f(c)=0$.

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

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

We want to determine the values of $k$ for which there is an $a$ such that $f(a)=0$.

From (a), if $k=-\frac{3}{2}, f(x)$ is constant and equal to $-\frac{1}{2}$, so has no roots.

Let $u=\sin ^{2} x$.

Then $u$ takes all values between 0 and 1 as $\sin x$ takes all values between -1 and 1 . Then we want to determine for which $k$ the equation

$$
(3+2 k) u^{2}-(3+2 k) u+(1+k)=0
$$

has a solution for $u$ with $0 \leq u \leq 1$.

First, we must ensure that the equation $(*)$ has real solutions, ie.

$$
\begin{aligned}
(3+2 k)^{2}-4(3+2 k)(1+k) & \geq 0 \\
(3+2 k)(3+2 k-4(1+k)) & \geq 0 \\
(3+2 k)(-1-2 k) & \geq 0 \\
(3+2 k)(1+2 k) & \leq 0
\end{aligned}
$$

This is true if and only if $-\frac{3}{2}<k \leq-\frac{1}{2}$. (We omit $k=-\frac{3}{2}$ because of the earlier comment.)

Next, we have to check for which values of $k$ the equation $(*)$ has a solution $u$ with $0 \leq u \leq 1$. We may assume that $-\frac{3}{2}<k \leq-\frac{1}{2}$.

To do this, we solve the equation $(*)$ using the quadratic formula to obtain

$$
u=\frac{(3+2 k) \pm \sqrt{(3+2 k)^{2}-4(3+2 k)(1+k)}}{2(3+2 k)}
$$

or

$$
u=\frac{(3+2 k) \pm \sqrt{-(3+2 k)(1+2 k)}}{2(3+2 k)}=\frac{1}{2} \pm \frac{1}{2} \sqrt{-\frac{1+2 k}{3+2 k}}
$$



Since $k>-\frac{3}{2}$ then $3+2 k>0$.

For $u$ to be between 0 and 1, we need to have

$$
0 \leq \sqrt{-\frac{1+2 k}{3+2 k}} \leq 1
$$

Thus

$$
0 \leq-\frac{1+2 k}{3+2 k} \leq 1
$$

Since $-\frac{3}{2}<k \leq-\frac{1}{2}$ then $3+2 k>0$ and $1+2 k \leq 0$, so the left inequality is true.

Therefore, we need $-\frac{1+2 k}{3+2 k} \leq 1$ or $-(1+2 k) \leq(3+2 k)$ (we can multiply by $(3+2 k)$ since it is positive), and so $-4 \leq 4 k$ or $k \geq-1$.

Combining with $-\frac{3}{2}<k \leq-\frac{1}{2}$ gives $-1 \leq k \leq-\frac{1}{2}$.
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

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

Now, we have

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

If we tried to solve $f(x)=0$, we would obtain

$$
(1+k)-\frac{1}{4}(3+2 k) \sin ^{2} 2 x=0
$$

or

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

(From (a), if $k=-\frac{3}{2}, f(x)$ is constant and equal to $-\frac{1}{2}$, so has no roots.)

In order to be able to solve this (first for $\sin 2 x$, then for $2 x$ then for $x$ ), we therefore need

$$
0 \leq \frac{4(1+k)}{3+2 k} \leq 1
$$

If $3+2 k>0$, we can multiply the inequality by $3+2 k$ to obtain

$$
0 \leq 4(1+k) \leq 3+2 k
$$

and so we get $k \geq-1$ from the left inequality and $k \leq-\frac{1}{2}$ from the right inequality.

Combining these with $-\frac{3}{2}<k$, we obtain $-1 \leq k \leq-\frac{1}{2}$.

If $3+2 k<0$, we would obtain $0 \geq 4(1+k) \geq 3+2 k$ which would give $k \leq-1$ and $k \geq-\frac{1}{2}$, which are inconsistent.

Therefore, $-1 \leq k \leq-\frac{1}{2}$.

Final answer: $[-1,-\frac{1}{2}]$