Calculate the area of the original triangle ABC.

The vertices are A(1,0), B(3,0), C(1,2).
The base of the triangle can be taken as the distance between A(1,0) and B(3,0) along the x-axis.
Base length $= 3 - 1 = 2$ units.
The height of the triangle is the perpendicular distance from C(1,2) to the x-axis.
Height $= 2$ units.
Area of triangle ABC $= \frac{1}{2} \times \text{base} \times \text{height}$

$$\text{Area}_{ABC} = \frac{1}{2} \times 2 \times 2 = 2 \text{ square units}$$

$$T = \begin{pmatrix} 3k & 1.6 \\ 3k & -0.9 \end{pmatrix}$$

The determinant of $T$ is $det(T) = (3k)(-0.9) - (1.6)(3k)$.

$$det(T) = -2.7k - 4.8k$$ $$det(T) = -7.5k$$

The area of the transformed triangle is given by the absolute value of the determinant of the transformation matrix multiplied by the area of the original triangle.

$$\text{Area}_{A'B'C'} = |det(T)| \times \text{Area}_{ABC}$$

Given $\text{Area}_{A'B'C'} = 6$ square units.

$$6 = |-7.5k| \times 2$$ $$6 = 7.5|k| \times 2$$ $$6 = 15|k|$$

$$|k| = \frac{6}{15}$$ $$|k| = \frac{2}{5}$$ $$|k| = 0.4$$

Therefore, $k$ can be 0.4 or $-0.4$.

$$\boxed{k = 0.4 \text{ or } k = -0.4}$$

9. Solve the equation $6 \cos^2 x + \sin x = 4$ for $0^\circ \le x \le 180^\circ$, giving the answer correct to 2 decimal places.

$$6(1 - \sin^2 x) + \sin x = 4$$ $$6 - 6 \sin^2 x + \sin x = 4$$

$$0 = 6 \sin^2 x - \sin x + 4 - 6$$ $$6 \sin^2 x - \sin x - 2 = 0$$

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-2)}}{2(6)}$$ $$y = \frac{1 \pm \sqrt{1 + 48}}{12}$$ $$y = \frac{1 \pm \sqrt{49}}{12}$$ $$y = \frac{1 \pm 7}{12}$$

This gives two possible values for $y$:

$$y_1 = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}$$ $$y_2 = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}$$

Case 1: $\sin x = \frac{2}{3}$
Since $\sin x$ is positive, $x$ is in the first or second quadrant.
The reference angle is $\arcsin\left(\frac{2}{3}\right)$.

$$x_1 = \arcsin\left(\frac{2}{3}\right) \approx 41.8103^\circ$$

To 2 decimal places, $x_1 = 41.81^\circ$.
The second quadrant solution is $180^\circ - x_1$:

$$x_2 = 180^\circ - 41.8103^\circ \approx 138.1897^\circ$$

To 2 decimal places, $x_2 = 138.19^\circ$.
Both $41.81^\circ$ and $138.19^\circ$ are within the range $0^\circ \le x \le 180^\circ$.

Case 2: $\sin x = -\frac{1}{2}$
Since $\sin x$ is negative, $x$ is in the third or fourth quadrant.
The reference angle is $\arcsin\left(\frac{1}{2}\right) = 30^\circ$.
Third quadrant solution: $180^\circ + 30^\circ = 210^\circ$.
Fourth quadrant solution: $360^\circ - 30^\circ = 330^\circ$.
Both $210^\circ$ and $330^\circ$ are outside the given range $0^\circ \le x \le 180^\circ$. Therefore, there are no solutions from this case.

The solutions for $x$ are:

$$\boxed{x = 41.81^\circ, 138.19^\circ}$$