Consider the following expression:

$$
E(x)=\sin ^{6} x+\cos ^{6} x+p\left(\sin ^{4} x+\cos ^{4} x\right)
$$

a) Determine the value of the parameter $p$ so that the value of the expression is independent of $x$.

b) For which values of $p$ does the equation $E(x)=0$ have (real) solutions?
Since $1=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}=\sin ^{6} x+\cos ^{6} x+3 \sin ^{2} x \cos ^{2} x$, and

$$
1=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x
$$

the expression $E(x)$ can be written in the following form:

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

If now the value of $E(x)$ is independent of $x$, then for example $E(0)=E(\pi / 4)$, that is, $(1+p)=(1+p)-(3+2 p) / 4$. Therefore, necessarily $p=-3 / 2$. Conversely, if $p=-3 / 2$, then the value of $E(x)$ is $-1 / 2$ for all $x$ according to (1). This answers the question a) of the problem.

As for b), we can assume that $p \neq-3 / 2$, since in that case $E(x)=0$ has no solution. According to (1), $E(x)=0$ if and only if

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

Such an $x$ exists if and only if

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

The first inequality holds if $p<-3 / 2$, or $p \geq-1$; the second inequality holds only for $-3 / 2<p \geq-1 / 2$. Thus, the equation $E(x)=0$ has a solution if and only if $-1 \leq p \leq-1 / 2$.