Solve the equation f(x)=0, complete the truth table for q p, and find alpha and alpha given alpha = (5)/(12).

We need to fill in the columns for $\neg p$ and $q \land \neg p$.

• For $\neg p$: This is the negation of $p$. If $p$ is True (T), $\neg p$ is False (F). If $p$ is False (F), $\neg p$ is True (T).
• For $q \land \neg p$: This is the conjunction of $q$ and $\neg p$. It is True only if both $q$ and $\neg p$ are True. Otherwise, it is False.

The completed table is:

$$\begin{array}{|c|c|c|c|} \hline p & q & \neg p & q \land \neg p \\ \hline \text{T} & \text{T} & \text{F} & \text{F} \\ \text{T} & \text{F} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{F} \\ \hline \end{array}$$

We use the trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$.

a) Find $\cos \alpha$
Substitute the given value of $\sin \alpha$:

$$\left(\frac{5}{12}\right)^2 + \cos^2 \alpha = 1$$ $$\frac{25}{144} + \cos^2 \alpha = 1$$

Subtract $\frac{25}{144}$ from both sides:

$$\cos^2 \alpha = 1 - \frac{25}{144}$$ $$\cos^2 \alpha = \frac{144}{144} - \frac{25}{144}$$ $$\cos^2 \alpha = \frac{119}{144}$$

Take the square root of both sides. Assuming $\alpha$ is an acute angle (or in a quadrant where cosine is positive):

$$\cos \alpha = \sqrt{\frac{119}{144}}$$ $$\cos \alpha = \frac{\sqrt{119}}{\sqrt{144}}$$ $$\cos \alpha = \boxed{\frac{\sqrt{119}}{12}}$$

b) Find $\tan \alpha$
Use the identity $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$:

$$\tan \alpha = \frac{\frac{5}{12}}{\frac{\sqrt{119}}{12}}$$

Multiply the numerator by the reciprocal of the denominator:

$$\tan \alpha = \frac{5}{12} \times \frac{12}{\sqrt{119}}$$ $$\tan \alpha = \frac{5}{\sqrt{119}}$$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{119}$:

$$\tan \alpha = \frac{5}{\sqrt{119}} \times \frac{\sqrt{119}}{\sqrt{119}}$$ $$\tan \alpha = \boxed{\frac{5\sqrt{119}}{119}}$$