Question 13 asks for the related acute angle for the solution of the equation $8 \tan x + 3 = -4$, accurate to two decimal places. Question 14 asks which expression is equivalent to $-\frac{\sqrt{3}}{2}$.

1. Problem Description

Question 13 asks for the related acute angle for the solution of the equation $8 \tan x + 3 = -4$, accurate to two decimal places.

Question 14 asks which expression is equivalent to $-\frac{\sqrt{3}}{2}$.

2. Solution Steps

Question 13:

First, we need to solve for $\tan x$.

$8 \tan x + 3 = -4$

$8 \tan x = -4 - 3$

$8 \tan x = -7$

$\tan x = -\frac{7}{8}$

Since $\tan x$ is negative, $x$ is in the second or fourth quadrant. To find the related acute angle, we ignore the negative sign and find the reference angle.

Let $\alpha$ be the related acute angle.

$\tan \alpha = \frac{7}{8}$

$\alpha = \arctan(\frac{7}{8})$

Using a calculator, $\alpha \approx 0.71$ radians.

Question 14:

We need to determine which of the given expressions is equal to $-\frac{\sqrt{3}}{2}$.

(1) $\sin(\frac{5\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$

(2) $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$

(3) $\sin(\frac{\pi}{6}) = \frac{1}{2}$

(4) $\cos(\frac{5\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$

Thus, $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$.

3. Final Answer

Question 13: 0.71

Question 14: $\cos(\frac{5\pi}{6})$