The problem states:

• $x \in [-\pi, 0]$
• $\cos(x) = \frac{7}{9}$

We are tasked with finding the exact value of $\tan(x)$.

Solution:

We use the Pythagorean identity:

$\sin^2(x) + \cos^2(x) = 1$

1. Substitute $\cos(x) = \frac{7}{9}$: $\sin^2(x) + \left(\frac{7}{9}\right)^2 = 1$

2. Compute $\cos^2(x)$: $\cos^2(x) = \left(\frac{7}{9}\right)^2 = \frac{49}{81}$ So: $\sin^2(x) = 1 - \frac{49}{81} = \frac{81}{81} - \frac{49}{81} = \frac{32}{81}$

3. Solve for $\sin(x)$: Since $x \in [-\pi, 0]$, $\sin(x) \leq 0$. Hence: $\sin(x) = -\sqrt{\frac{32}{81}} = -\frac{\sqrt{32}}{9} = -\frac{4\sqrt{2}}{9}$

4. Calculate $\tan(x)$: Using $\tan(x) = \frac{\sin(x)}{\cos(x)}$: $\tan(x) = \frac{-\frac{4\sqrt{2}}{9}}{\frac{7}{9}} = -\frac{4\sqrt{2}}{7}$

Final Answer:

$\tan(x) = -\frac{4\sqrt{2}}{7}$

Let me know if you'd like further details or clarifications!

Related Questions:

1. What is the geometric interpretation of $\tan(x)$ on the unit circle?
2. How does the quadrant of $x$ influence the signs of trigonometric functions?
3. Derive $\sin(x)$ and $\cos(x)$ using the Pythagorean theorem for a general case.
4. What is the exact value of $\sec(x)$ in this case?
5. How would the solution change if $x \in [0, \pi]$?

Tip:

Remember to always check the quadrant of $x$ to correctly determine the signs of $\sin(x)$ and $\tan(x)$!