(a) Expand and simplify fully the expression $(a-1)\left(6 a^{2}-a-1\right)$.

(b) Determine all values of $\theta$ with $6 \cos ^{3} \theta-7 \cos ^{2} \theta+1=0$ and $-180^{\circ}<\theta<180^{\circ}$.

Round each answer to 1 decimal place where appropriate.

(Note that $\cos ^{3} \theta=(\cos \theta)^{3}$.)

(c) Determine all values of $\theta$ with $6 \cos ^{3} \theta-7 \cos ^{2} \theta+1<0$ and $-180^{\circ}<\theta<180^{\circ}$.
(a) Expanding and simplifying, we obtain

$$
(a-1)\left(6 a^{2}-a-1\right)=6 a^{3}-a^{2}-a-6 a^{2}+a+1=6 a^{3}-7 a^{2}+1
$$

(b) To solve the equation $6 \cos ^{3} \theta-7 \cos ^{2} \theta+1=0$, we first make the substitution $a=\cos \theta$. The equation becomes $6 a^{3}-7 a^{2}+1=0$.

From (a), factoring the left side gives the equation $(a-1)\left(6 a^{2}-a-1\right)=0$.

We further factor $6 a^{2}-a-1$ as $(3 a+1)(2 a-1)$.

This gives $(a-1)(3 a+1)(2 a-1)=0$.

Thus, $a=1$ or $a=-\frac{1}{3}$ or $a=\frac{1}{2}$.

This tells us that the solutions to the original equation are the values of $\theta$ in the range $-180^{\circ}<\theta<180^{\circ}$ with $\cos \theta=1$ or $\cos \theta=-\frac{1}{3}$ or $\cos \theta=\frac{1}{2}$.

If $-180^{\circ}<\theta<180^{\circ}$ and $\cos \theta=1$, then $\theta=0^{\circ}$.

If $-180^{\circ}<\theta<180^{\circ}$ and $\cos \theta=-\frac{1}{3}$, then $\theta \approx 109.5^{\circ}$ or $\theta \approx-109.5^{\circ}$. (The positive value for $\theta$ can be obtained using a calculator. The negative value can be obtained by thinking about $\cos \theta$ as an even function of $\theta$ or by picturing either the graph of $y=\cos \theta$ or the unit circle.)

If $-180^{\circ}<\theta<180^{\circ}$ and $\cos \theta=\frac{1}{2}$, then $\theta=60^{\circ}$ or $\theta=-60^{\circ}$.

Therefore, the solutions to the equation $6 \cos ^{3} \theta-7 \cos ^{2} \theta+1=0$, rounded to one decimal place as appropriate, are $0^{\circ}, 60^{\circ},-60^{\circ}, 109.5^{\circ},-109.5^{\circ}$.

(c) To solve the inequality $6 \cos ^{3} \theta-7 \cos ^{2} \theta+1<0$, we factor the left side to obtain

$$
(\cos \theta-1)(3 \cos \theta+1)(2 \cos \theta-1)<0
$$

From (b), we know the values of $\theta$ at which the left side equals 0 , so we examine the intervals between these values and look at the sign (positive or negative) of each of the factors in these intervals. We complete the final column of the table below by noting that the product of three positive numbers is positive, the product of two positives with one negative is negative, the product of one positive and two negatives is positive, and the product of three negatives is negative.

| Range of $\theta$ | Range of $\cos \theta$ | $\cos \theta-1$ | $3 \cos \theta+1$ | $2 \cos \theta-1$ | Product |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $-180^{\circ}<\theta<-109.5^{\circ}$ | $-1<\cos \theta<-\frac{1}{3}$ | - | - | - | - |
| $-109.5^{\circ}<\theta<-60^{\circ}$ | $-\frac{1}{3}<\cos \theta<\frac{1}{2}$ | - | + | - | + |
| $-60^{\circ}<\theta<0^{\circ}$ | $\frac{1}{2}<\cos \theta<1$ | - | + | + | - |
| $0^{\circ}<\theta<60^{\circ}$ | $\frac{1}{2}<\cos \theta<1$ | - | + | + | - |
| $60^{\circ}<\theta<109.5^{\circ}$ | $-\frac{1}{3}<\cos \theta<\frac{1}{2}$ | - | + | - | + |
| $109.5^{\circ}<\theta<180^{\circ}$ | $-1<\cos \theta<-\frac{1}{3}$ | - | - | - | - |

From this analysis, the values of $\theta$ for which $6 \cos ^{3} \theta-7 \cos ^{2} \theta+1<0$ are

$-180^{\circ}<\theta<-109.5^{\circ}$ and $-60^{\circ}<\theta<0^{\circ}$ and $0^{\circ}<\theta<60^{\circ}$ and $109.5^{\circ}<\theta<180^{\circ}$

We could also have determined these intervals by looking only at positive values for $\theta$ and then using the fact that $\cos \theta$ is an even function to determine the negative values.