The function \(y = \sin(a(x + b))\) is a sine wave with an amplitude of 1. The period of the sine function is given by \(\frac{2\pi}{a}\). From the graph, the period is \(4\pi\), so:

\(\frac{2\pi}{a} = 4\pi\)

Solving for \(a\), we get:

\(a = \frac{1}{2}\)

The phase shift is determined by \(b\). The graph starts at \(x = -\frac{\pi}{3}\), which corresponds to the phase shift. Therefore, \(b = \frac{\pi}{3}\).

The equation of the curve is given by \(y = p \sin(q\theta) + r\).

(a) The amplitude of the sine function is the coefficient \(p\). From the graph, the maximum value is 1 and the minimum value is -5, giving an amplitude of \(\frac{1 - (-5)}{2} = 3\). Thus, \(p = 3\).

(b) The period of the sine function is given by \(\frac{2\pi}{q}\). From the graph, the period is \(4\pi\), so \(\frac{2\pi}{q} = 4\pi\). Solving for \(q\), we get \(q = \frac{1}{2}\).

(c) The vertical shift \(r\) is the average of the maximum and minimum values of the function. Thus, \(r = \frac{1 + (-5)}{2} = -2\).

(a) The graph of \(y = a \cos(bx) + c\) shows a cosine wave with a maximum value of 8 and a minimum value of -2. The amplitude \(a\) is half the distance between the maximum and minimum values, so \(a = \frac{8 - (-2)}{2} = 5\).

The period of the cosine function is \(\frac{2\pi}{b}\). From the graph, the period is \(\pi\), so \(\frac{2\pi}{b} = \pi\), giving \(b = 2\).

The vertical shift \(c\) is the average of the maximum and minimum values, so \(c = \frac{8 + (-2)}{2} = 3\).

(b)(i) For \(a \cos(bx) + c = \frac{6}{\pi} x\), substitute \(a = 5\), \(b = 2\), \(c = 3\) to get \(5 \cos(2x) + 3 = \frac{6}{\pi} x\). The number of solutions is the number of intersections of the line \(y = \frac{6}{\pi} x\) with the cosine graph in the interval \(0 \leq x \leq 2\pi\), which is 3.

(b)(ii) For \(a \cos(bx) + c = 6 - \frac{6}{\pi} x\), substitute \(a = 5\), \(b = 2\), \(c = 3\) to get \(5 \cos(2x) + 3 = 6 - \frac{6}{\pi} x\). The number of solutions is the number of intersections of the line \(y = 6 - \frac{6}{\pi} x\) with the cosine graph in the interval \(0 \leq x \leq 2\pi\), which is 2.

The graph shown is a transformation of the basic tangent function \(y = \tan x\).

The vertical stretch factor \(a\) is determined by the amplitude of the graph. The graph reaches approximately 3 at \(x = \frac{\pi}{2}\), indicating \(a = 2\).

The horizontal shift \(b\) is determined by the phase shift of the graph. The graph crosses the x-axis at \(x = \frac{\pi}{4}\), so \(b = \frac{\pi}{4}\).

The vertical shift \(c\) is determined by the midline of the graph. The midline appears to be at \(y = 1\), so \(c = 1\).

The graph of \(y = a + b \sin x\) is shown with a maximum value of 1 and a mid value of -2, which indicates \(a = -2\).

The amplitude \(b\) is the distance from the midline to the maximum or minimum. Since the maximum is 1 and the midline is -2, the amplitude is \(1 - (-2) = 3\).

Thus, \(a = -2\) and \(b = 3\).

The graph of \(y = a + b \sin x\) is a vertically shifted sine wave.

The midline of the sine wave is at \(y = 1\), which indicates that \(a = 1\).

The amplitude of the sine wave is the distance from the midline to the maximum or minimum point. The maximum value is \(3\) and the minimum value is \(-1\), so the amplitude is \(\frac{3 - 1}{2} = 2\).

Thus, \(b = 2\).

(i) The amplitude of the sine function is the distance from the midline to the maximum, which is \(9 - 3 = 6\). Therefore, \(a = 6\).

The period of the sine function is \(2\pi\) divided by the coefficient of \(x\), which is \(b\). The period is \(\pi\), so \(b = 2\).

The vertical shift \(c\) is the midline of the graph, which is at \(y = 3\). Therefore, \(c = 3\).

(ii) To find the smallest \(x\) for which \(y = 0\), set the equation to zero: \(6\sin(2x) + 3 = 0\).

Solving for \(\sin(2x)\):

\(6\sin(2x) = -3\)

\(\sin(2x) = -\frac{1}{2}\)

The general solution for \(\sin(\theta) = -\frac{1}{2}\) is \(\theta = \frac{7\pi}{6} + 2k\pi\) or \(\theta = \frac{11\pi}{6} + 2k\pi\).

For \(2x = \frac{7\pi}{6}\), solve for \(x\):

\(x = \frac{7\pi}{12}\)

Converting to decimal gives \(x \approx 1.83\).

(a) The function \(y = 3 \cos 2x + 2\) is a cosine function scaled and shifted. The cosine function \(\cos 2x\) has a range of \([-1, 1]\). Therefore, \(3 \cos 2x\) has a range of \([-3, 3]\). Adding 2 shifts this range to \([-1, 5]\). Thus, the greatest value is 5 and the least value is -1.

(b) The graph of \(y = 3 \cos 2x + 2\) is a cosine wave starting at its maximum value of 5 at \(x = 0\), decreasing to its minimum value of -1 at \(x = \frac{\pi}{2}\), and returning to its maximum value of 5 at \(x = \pi\). This completes one full cycle.

(c)(i) For \(k = -3\), the line \(y = -3x\) does not intersect the curve \(y = 3 \cos 2x + 2\) within the given range, so there are 0 solutions.

(c)(ii) For \(k = 1\), the line \(y = x\) intersects the curve \(y = 3 \cos 2x + 2\) at two points within the given range, so there are 2 solutions.

(c)(iii) For \(k = 3\), the line \(y = 3x\) intersects the curve \(y = 3 \cos 2x + 2\) at one point within the given range, so there is 1 solution.

(i) The function is given by \(f(x) = a + b \cos 2x\). We have two conditions:

\(f(0) = a + b \cos 0 = -1\) which simplifies to \(a + b = -1\).

\(f\left(\frac{1}{2}\pi\right) = a + b \cos \pi = 7\) which simplifies to \(a - b = 7\).

Solving these two equations:

\(a + b = -1\)

\(a - b = 7\)

Add the equations: \(2a = 6\) so \(a = 3\).

Substitute \(a = 3\) into \(a + b = -1\):

\(3 + b = -1\) so \(b = -4\).

Thus, \(a = 3\) and \(b = -4\).

(ii) To find the x-coordinates where the curve intersects the x-axis, set \(f(x) = 0\):

\(3 - 4 \cos 2x = 0\)

\(\cos 2x = \frac{3}{4}\)

Solving for \(x\), we find:

\(2x = \cos^{-1}\left(\frac{3}{4}\right)\)

\(x = \frac{1}{2} \cos^{-1}\left(\frac{3}{4}\right)\)

Calculating gives \(x \approx 0.36\) and \(x \approx 2.78\).

(iii) The graph of \(y = f(x)\) is a cosine wave starting at \(y = -1\) when \(x = 0\) and reaching a maximum of \(y = 7\) at \(x = \frac{1}{2}\pi\), then returning to \(y = -1\) at \(x = \pi\). The graph oscillates once between \(x = 0\) and \(x = \pi\).

(i) To sketch the graphs of \(y = 2 \sin x\) and \(y = \cos 2x\) for \(0 \leq x \leq \pi\):

- The graph of \(y = 2 \sin x\) will have a maximum value of 2 and a minimum value of 0 within the interval. It completes half a cycle from 0 to \(\pi\).

- The graph of \(y = \cos 2x\) completes a full cycle from 0 to \(\pi\), oscillating between 1 and -1.

(ii) The number of solutions to \(2 \sin x = \cos 2x\) corresponds to the number of intersection points of the two graphs within the interval \(0 \leq x \leq \pi\). From the sketch, there are 2 points of intersection.

(i) The graph of \(y = 3 \sin x\) is a sine wave with amplitude 3, oscillating between -3 and 3, over the interval \(-\pi \leq x \leq \pi\). The maximum point is at \(x = \frac{\pi}{2}\), where \(y = 3\).

(ii) The line \(y = kx\) passes through the maximum point \(\left( \frac{\pi}{2}, 3 \right)\). Substituting into the line equation gives:

\(3 = k \left( \frac{\pi}{2} \right)\)

Solving for \(k\):

\(k = \frac{6}{\pi}\)

(iii) To find the other intersection point, set \(y = 3 \sin x = kx\) and solve for \(x\):

\(3 \sin x = \frac{6}{\pi} x\)

At \(x = -\frac{\pi}{2}\), \(\sin x = -1\), so:

\(3(-1) = \frac{6}{\pi} \left(-\frac{\pi}{2}\right)\)

\(-3 = -3\)

Thus, the coordinates are \(\left( -\frac{\pi}{2}, -3 \right)\).

(i) We have two equations from the given conditions:

\(a \sin\left(\frac{1}{2}\pi\right) + b = 2\)

\(a \cdot 1 + b = 2\)

\(a + b = 2\)

\(a \sin\left(\frac{3}{2}\pi\right) + b = -8\)

\(a \cdot (-1) + b = -8\)

\(-a + b = -8\)

Solving these equations:

\(a + b = 2\)

\(-a + b = -8\)

Add the equations:

\(2b = -6\)

\(b = -3\)

Substitute \(b = -3\) into \(a + b = 2\):

\(a - 3 = 2\)

\(a = 5\)

(ii) To find \(x\) for which \(f(x) = 0\):

\(5 \sin x - 3 = 0\)

\(5 \sin x = 3\)

\(\sin x = \frac{3}{5}\)

\(x = \sin^{-1}\left(\frac{3}{5}\right) \approx 0.64\)

Since \(\sin x\) is positive in the first and second quadrants:

\(x = \pi - 0.64 \approx 2.50\)

(iii) The graph of \(y = 5 \sin x - 3\) is a sine wave with amplitude 5, shifted down by 3 units. It starts at \(y = -3\) when \(x = 0\), reaches a maximum of \(y = 2\) at \(x = \frac{1}{2}\pi\), and a minimum of \(y = -8\) at \(x = \frac{3}{2}\pi\).

(i) To solve \(fg(x) = 1\), substitute \(g(x) = \frac{1}{2}x\) into \(f(x)\):

\(fg(x) = 2 - 3 \cos\left(\frac{1}{2}x\right) = 1\)

\(2 - 3 \cos\left(\frac{1}{2}x\right) = 1 \rightarrow \cos\left(\frac{1}{2}x\right) = \frac{1}{3}\)

\(\frac{1}{2}x = \cos^{-1}\left(\frac{1}{3}\right)\)

\(x = 2 \times \cos^{-1}\left(\frac{1}{3}\right) \approx 2.46\)

Alternative: Solve \(f(y) = 1 \rightarrow y = 1.23 \rightarrow \frac{1}{2}x = 1.23 \rightarrow x = 2.46\)

(ii) The graph of \(y = f(x) = 2 - 3 \cos x\) is a cosine curve. Sketch one cycle from \(0\) to \(2\pi\), starting and ending at the same negative \(y\) value, with a peak at \(y = 5\) and a trough at \(y = -1\).

(i) Start with the equation \(2 \cos x + 3 \sin x = 0\).

Rearrange to \(\tan x = -\frac{2}{3}\).

Find \(x\) using the inverse tangent function: \(x = 146.3^\circ\) or \(x = 326.3^\circ\).

(ii) Sketch the graphs of \(y = 2 \cos x\) and \(y = -3 \sin x\) over the interval \(0^\circ \leq x \leq 360^\circ\). The cosine graph starts at the top of the curve, while the sine graph is inverted and starts on the x-axis.

(iii) From the graphs, determine where \(2 \cos x + 3 \sin x > 0\). This occurs when \(x < 146.3^\circ\) and \(x > 326.3^\circ\).

(i) The curve \(y = \sin 2x\) completes two cycles from \(0\) to \(2\pi\), starting and finishing on the x-axis, with a maximum of 1 and a minimum of -1. The curve \(y = \cos x - 1\) completes one cycle, starting and finishing on the x-axis, with a minimum point at -2.

(ii) (a) For \(2 \sin 2x + 1 = 0\), we have \(\sin 2x = -\frac{1}{2}\). The solutions for \(\sin 2x = -\frac{1}{2}\) in the interval \(0 \leq x \leq 2\pi\) are four points where the sine curve intersects the line \(y = -\frac{1}{2}\).

(b) For \(\sin 2x - \cos x + 1 = 0\), the solutions are the points where the curves \(y = \sin 2x\) and \(y = \cos x - 1\) intersect. There are three such intersections in the interval \(0 \leq x \leq 2\pi\).

(i) The graph of \(y = \sin x\) starts at \((0,0)\), peaks at \((90^\circ, 1)\), and returns to \((180^\circ, 0)\). The graph of \(y = \cos 2x\) starts at \((0, 1)\), completes one full cycle by \(180^\circ\), and oscillates between -1 and 1.

(ii) To verify \(x = 30^\circ\) is a root, calculate \(\sin 30^\circ = 0.5\) and \(\cos 60^\circ = 0.5\), confirming \(\sin 30^\circ = \cos 60^\circ\). The other root is found by solving \(\sin x = \cos 2x\) for \(0^\circ \leq x \leq 180^\circ\), giving \(x = 150^\circ\).

(iii) The inequality \(\sin x < \cos 2x\) holds for \(0 \leq x < 30\) and \(150 < x \leq 180\).

(i) The graph of \(y = \cos 2\theta\) oscillates between -1 and 1 and completes two full oscillations in the interval \([0, 2\pi]\). The line \(y = \frac{1}{2}\) is a horizontal line. The intersections of these two graphs represent the solutions to the equation \(2\cos 2\theta - 1 = 0\).

(ii) The equation \(2\cos 2\theta - 1 = 0\) simplifies to \(\cos 2\theta = \frac{1}{2}\). The solutions for \(2\theta\) are \(60^\circ, 300^\circ\) (or \(\frac{\pi}{3}, \frac{5\pi}{3}\)) within one period \([0, 2\pi]\). Since \(2\theta\) ranges from \(0\) to \(4\pi\), there are 4 solutions: \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

(iii) The interval \(10\pi \leq \theta \leq 20\pi\) is 5 times the length of \(0 \leq \theta \leq 2\pi\). Therefore, the number of roots is 5 times the number of roots in the interval \(0 \leq \theta \leq 2\pi\), which is \(5 \times 4 = 20\).

(i) The curve \(y = 2 \sin x\) is a sine wave with amplitude 2, period \(2\pi\), and it oscillates between -2 and 2. The curve completes one full cycle from \(0\) to \(2\pi\).

(ii) To find the number of real roots of the equation \(2\pi \sin x = \pi - x\), we rearrange it to \(2\pi \sin x + x = \pi\). This can be rewritten as \(y = 2\pi \sin x\) and \(y = \pi - x\). The line \(y = \pi - x\) can be rewritten as \(y = 1 - \frac{x}{\pi}\) by dividing through by \(\pi\).

The line \(y = 1 - \frac{x}{\pi}\) passes through the points \((0, 1)\) and \((\pi, 0)\). By sketching both the sine curve and the line, we observe that they intersect at three points within the interval \(0 \leq x \leq 2\pi\), indicating there are 3 real roots.

The curve \(y = 3 \cos 2x\) is a cosine function with amplitude 3 and period \(\pi\). It completes one full oscillation from \(x = 0\) to \(x = \pi\). The range of the curve is from -3 to 3.

The line \(x + 2y = \pi\) can be rewritten as \(y = \frac{\pi - x}{2}\). This is a straight line with a negative slope of -1/2 and a y-intercept of \(\frac{\pi}{2}\).

To sketch the curve, plot the cosine wave starting at \(y = 3\) when \(x = 0\), decreasing to \(y = -3\) at \(x = \frac{\pi}{2}\), and returning to \(y = 3\) at \(x = \pi\).

To sketch the line, plot the intercepts: when \(x = 0\), \(y = \frac{\pi}{2}\); when \(y = 0\), \(x = \pi\). Draw the line through these points.

(i) Given the maximum value of \(f(x) = a - b \cos x\) is 10, and the minimum value is \(-2\), we have:

\(a + b = 10\) (when \(\cos x = -1\))

\(a - b = -2\) (when \(\cos x = 1\))

Solving these equations simultaneously:

\(a + b = 10\)

\(a - b = -2\)

Adding the equations gives:

\(2a = 8 \Rightarrow a = 4\)

Substituting \(a = 4\) into \(a + b = 10\):

\(4 + b = 10 \Rightarrow b = 6\)

(ii) To solve \(f(x) = 0\):

\(4 - 6 \cos x = 0\)

\(\cos x = \frac{2}{3}\)

Using the inverse cosine function, \(x = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\) or \(360^\circ - 48.2^\circ = 311.8^\circ\).

(iii) The sketch of \(y = f(x)\) should show one cycle of a cosine wave, starting at \(-2\), peaking at 10, and returning to \(-2\) at \(x = 360^\circ\).

The original stationary point is at \((p, q)\).

The transformation applied to the curve is \(y = -3f\left(\frac{1}{4}(x + 8)\right)\).

This transformation involves:

• A horizontal translation of \(-8\) units (due to \(x + 8\)).
• A horizontal scaling by a factor of \(4\) (due to \(\frac{1}{4}x\)).
• A vertical scaling by a factor of \(-3\) (due to \(-3f(x)\)).

For the x-coordinate:

Start with the horizontal translation: \(x' = x + 8\), so \(x = x' - 8\).

Apply the horizontal scaling: \(x'' = 4x\), so \(x'' = 4(x' - 8) = 4p - 8\).

For the y-coordinate:

Apply the vertical scaling: \(y'' = -3y\), so \(y'' = -3q\).

Thus, the coordinates of the stationary point for the transformed curve are \((4p - 8, -3q)\).

To transform the graph of \(y = \sin x\) to \(y = f(x) = 3 + 2 \sin \frac{1}{4}x\), apply the following transformations in order:

1. Stretch the graph horizontally by a factor of 4 in the x-direction. This changes \(\sin x\) to \(\sin \frac{1}{4}x\).

2. Stretch the graph vertically by a factor of 2 in the y-direction. This changes \(\sin \frac{1}{4}x\) to \(2 \sin \frac{1}{4}x\).

3. Translate the graph vertically by 3 units upwards. This changes \(2 \sin \frac{1}{4}x\) to \(3 + 2 \sin \frac{1}{4}x\).

1. Translate the graph by \(\begin{pmatrix} 30^\circ \\ 0 \end{pmatrix}\) to account for the phase shift of \(-30^\circ\).

2. Stretch the graph by a factor of 2 in the x-direction to account for the coefficient \(\frac{1}{2}\) in front of \(x\).

3. Stretch the graph by a factor of 4 in the y-direction to account for the amplitude change from 1 to 4.

The original curve is \(y = \sin 2x - 5x\).

First, reflect the curve in the \(y\)-axis. This changes \(x\) to \(-x\), so the equation becomes \(y = \sin(-2x) - 5(-x) = -\sin 2x + 5x\).

Next, stretch the curve by a scale factor of \(\frac{1}{3}\) in the \(x\)-direction. This changes \(x\) to \(\frac{x}{\frac{1}{3}} = 3x\), so the equation becomes \(y = -\sin(2 \cdot 3x) + 5(3x) = -\sin 6x + 15x\).

(d) To map \(y = f(x)\) to \(y = g(x)\):

1. First, apply a horizontal stretch by a scale factor of \(\frac{1}{2}\) because \(g(x) = f(2x)\).
2. Then, translate the graph vertically by 4 units upwards, as indicated by \(g(x) = f(2x) + 4\).

(e) To map \(y = f(x)\) to \(y = h(x)\):

1. First, translate the graph horizontally by \(-\frac{\pi}{2}\) units, as indicated by \(h(x) = 2f\left(x + \frac{1}{2}\pi\right)\).
2. Then, apply a vertical stretch by a scale factor of 2, as indicated by the multiplication by 2 in \(h(x) = 2f\left(x + \frac{1}{2}\pi\right)\).

The transformation applied to the lower curve \(y = \\cos \\theta\) to obtain the upper curve involves both a vertical stretch and a horizontal stretch.

The vertical stretch is by a factor of 2, and the horizontal stretch is by a factor of 2, which affects the period of the cosine function. Additionally, there is a vertical translation upwards by 3 units.

Thus, the equation of the upper curve is:

\(y = 3 + 2 \cos \left( \frac{1}{2} \theta \right)\)

(a) The function \(f(x) = 2 - 3 \sin 2x\) has a range determined by the values of \(\sin 2x\), which varies from \(-1\) to \(1\). Therefore, \(f(x)\) ranges from \(2 - 3(1) = -1\) to \(2 - 3(-1) = 5\).

The function \(g(x) = -2f(x)\) will have a range that is twice the range of \(f(x)\) and inverted. Thus, \(g(x)\) ranges from \(-2(5) = -10\) to \(-2(-1) = 2\).

(b) The graph of \(y = g(x)\) is a reflection of \(y = f(x)\) in the x-axis and stretched by a factor of 2 in the y direction.

(c) To map \(y = f(x)\) onto \(y = h(x)\), perform the following transformations:

• Reflect in the x-axis.
• Stretch by a factor of 2 in the y direction.
• Translate by \(-\pi\) in the x direction, or equivalently, translate by \(\begin{pmatrix} 0 \\ -\pi \end{pmatrix}\).

(a) The function \(f(x) = \frac{3}{2} \cos 2x + \frac{1}{2}\) has a cosine component \(\frac{3}{2} \cos 2x\) which varies between \(-\frac{3}{2}\) and \(\frac{3}{2}\). Adding \(\frac{1}{2}\) shifts this range to \(-1\) to \(2\). Thus, the range of \(f(x)\) is \(-1 < f(x) < 2\).

(b) Since the x-axis is a tangent to \(y = g(x)\), the minimum value of \(g(x)\) must be 0. Therefore, \(f(x) + k = 0\) when \(f(x)\) is at its minimum, which is \(-1\). Solving \(-1 + k = 0\) gives \(k = 1\). The transformation is a translation by 1 unit upwards parallel to the y-axis.

(c) The reflection of \(y = f(x)\) in the x-axis is given by \(y = -f(x)\). Therefore, \(y = -\left( \frac{3}{2} \cos 2x + \frac{1}{2} \right) = -\frac{3}{2} \cos 2x - \frac{1}{2}\).