Detailed Answer

a) To solve this, we need \( \cos(x) = \sin(3x) \). The value where sin and cos are equal is where they are both \( \frac{\sqrt{2}}{2} \). For cosine, the first time this occurs is at \( \frac{\pi}{4} \), and since sin is positive in the first and second quadrants, it will also be \( \frac{\sqrt{2}}{2} \) at \( \frac{\pi}{4} \) and \( \frac{3\pi}{4} \). This works out as we have \( \sin(3x) \).

b) To find the area of the region, we need to set up an integral:

\[ \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \left(\sin(3x) - \cos(x)\right) \, dx = \left[ -\frac{\cos(3x)}{3} - \sin(x) \right]_{\frac{\pi}{8}}^{\frac{\pi}{4}} \]

\[ = -\frac{\cos\left(\frac{3\pi}{4}\right)}{3} - \sin\left(\frac{\pi}{4}\right) - \left(-\frac{\cos\left(\frac{3\pi}{8}\right)}{3} - \sin\left(\frac{\pi}{8}\right)\right) \]

\[ = \frac{-\frac{\sqrt{2}}{2}}{3} - \frac{\sqrt{2}}{2} + \frac{\cos\left(\frac{3\pi}{8}\right)}{3} + \sin\left(\frac{\pi}{8}\right) \]

\[ = -\frac{\sqrt{2}}{3} + \frac{\cos\left(\frac{3\pi}{8}\right)}{3} + \sin\left(\frac{\pi}{8}\right) \]

\[ = \frac{\cos\left(\frac{3\pi}{8}\right) - \sqrt{2}}{3} + \sin\left(\frac{\pi}{8}\right) \]

Detailed Answer

a) First, we need to take the arctangent of both sides:

\[ 5x + 6 = \arctan(\sqrt{3}) = \frac{\pi}{3} \]

\[ x = \frac{\frac{\pi}{3} - 6}{5} = \frac{\pi - 18}{15} \]

Detailed Answer

a) From the graph, it can be seen that the distance between the maximum and the center line is 5. Hence, the amplitude is 5. The graph is flipped around the x-axis, so \(a = -5\).

b) The period can be read off the graph as \(\pi\).

c) The period of this function can be interpreted as:

\[ T = \frac{2\pi}{b} = \pi \quad \Rightarrow \quad b = 2 \]

d) Evaluating \(f\left(\frac{\pi}{3}\right)\):

\[ f\left(\frac{\pi}{3}\right) = -5\cos{\left(2 \cdot \frac{\pi}{3}\right)} = -2.5 \]

Detailed Answer

\[\sin{(x)} \cdot \frac{\sin{(x)}}{\cos{(x)}} - \sin^2{(x)} = 0 \]

\[ \sin^2{x} \left( \frac{1}{\cos{(x)}} - 1 \right) = 0 \]

\[ \sin^2{x} = 0 \]

\[ \frac{1}{\cos{x}} = 1 \]

From these two equations, we get:

\[ x = -\pi, 0, \pi \]

Detailed Answer

a) For this, it helps to draw a right-angled triangle and label one of the angles \( x \). It is known that the tangent of this angle is 2, therefore we can say that the side opposite to \( x \) is 2, and the side adjacent to it is 1.

Using Pythagoras, we determine the hypotenuse of the triangle:

\[ \text{Hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{5} \]

From this triangle, we know:

\[ \sin{(x)} = \frac{2}{\sqrt{5}}, \quad \cos{(x)} = \frac{1}{\sqrt{5}} \]

Then:

\[ \sin{(2x)} = 2\sin{(x)}\cos{(x)} = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5} \]

Detailed Answer

a) The first time \(\sin(x) = \frac{1}{2}\) is at \(\frac{\pi}{6}\). Thus, we need:

\[ 2x + \frac{\pi}{3} = \frac{\pi}{6} \]

However, solving this gives:

\[ x = -\frac{\pi}{12} \]

This value of \(x\) is negative, but we need \(x\) to be positive. The next value where \(\sin(x) = \frac{1}{2}\) is at \(\frac{5\pi}{6}\). Substituting, we get:

\[ 2x + \frac{\pi}{3} = \frac{5\pi}{6} \]

\[ x = \frac{\pi}{4} \]

Detailed Answer

a) We need to use trigonometric identities to rewrite \(\cos{(2x)}\):

\[ \cos{(2x)} + 3\cos{(x)} - 1 = 2\cos^2{x} - 1 + 3\cos{(x)} - 1 = 2\cos^2{x} + 3\cos{(x)} - 2 \]

\[ = 2\cos^2{x} + 4\cos{(x)} - \cos{(x)} - 2 = 2\cos{(x)}(\cos{(x)} + 2) - (\cos{(x)} + 2) \]

\[ = (2\cos{(x)} - 1)(\cos{(x)} + 2) \]

b) We now need to solve the equation:

\[ (2\cos{(x)} - 1)(\cos{(x)} + 2) = 0\]

This product can be \(0\) when one or both of the terms are \(0\):

\[ \cos{(x)} = -2 \;\; \text{has no solutions} \]

\[ \cos{(x)} = \frac{1}{2} \;\; \text{when } x = \frac{\pi}{6} \text{ or } \frac{11\pi}{6} \]

To get all solutions, we must add \(2k\pi\) to all answers here to cover all possible values of \(x\).

Detailed Answer

a) We need to find the pressures when \(t = 0\):

\[ p_a(0) = \sin\left(0.6(0) + 20\right) - (0) + 27 = 27.9 \]

\[ p_b(0) = \sin\left(0.6(0) + 20\right) - 0.5(0) + 20 = 20.9 \]

b) For this, we can plot the two graphs in our GDC and find their intersection point, which is \((14, 12.9)\).

c) First, we need to find the initial rate of change of the pressure in \(A\), which can be done by taking the derivative and then plugging in \(t = 0\). However, it is easier if we use the \(p_b\) function we have plotted in the previous part and use the analysis function of the GDC to find the derivative when \(t = 0\). This will be \(-0.755\).

We can then plot the derivative function of \(B\), \(p'_b\), and find its intersection with the line \(y = -0.75515\). The first value of \(t\) where it intersects is at \(1.43\).

Detailed Answer

a)

\[ P(X=210) = 0.0147 \]

b)

\[ P(X>240) = 0.212 \]

c) It will be a binomial distribution with the probability obtained in part (b):

\[ P(X<4) = 0.910 \]

d) We have to first calculate the probability of a cheeseburger to weigh more than 240g:

\[ P(X>240) = 0.356 \]

There will be two possibilites of a burger to weigh more than 240g, it can either be a cheeseburger or a hamburger, so the probabilities have to be added:

\[ 0.55 \times 0.356 + 0.45 \times 0.212 = 0.291\]

e)

\[ P(C|>240) = \frac{P(C \ \cap >240)}{P(>240)} = \frac{0.55 \times 0.356}{0.291} = 0.672 \]