(a) To sketch the graph of \(y = f^{-1}(x)\), reflect the graph of \(y = f(x)\) across the line \(y = x\). Ensure the line \(y = x\) is drawn correctly and the inverse graph reaches \(y = 2\pi\) and \(x\)-axis, crossing the line \(y = x\) in the correct squares.

(b) Start with \(y = 3 + 2 \sin \frac{1}{4}x\). Rearrange to find \(\sin \frac{1}{4}x = \frac{y-3}{2}\). Solve for \(x\) to get \(x = 4 \sin^{-1} \left( \frac{y-3}{2} \right)\). Thus, \(f^{-1}(x) = 4 \sin^{-1} \left( \frac{x-3}{2} \right)\).

(c) Extend the graph of \(g(x)\) from \(y\)-intercept to \(-2\pi\). The graph should start to level off as \(x \to -2\pi\). The function \(g(x)\) has an inverse because it is always increasing, passing the horizontal line test.

(i) (a) For \(k = 2\), \(f(x) = 3 - 4 \cos^2 x\). The maximum value of \(\cos^2 x\) is 1, and the minimum is 0. Thus, the range of \(f(x)\) is from \(3 - 4(1) = -1\) to \(3 - 4(0) = 3\). Therefore, the range is \([-1, 3]\).

(b) To solve \(f(x) = 1\), set \(3 - 4 \cos^2 x = 1\). This simplifies to \(\cos^2 x = \frac{1}{2}\), giving \(\cos x = \pm \frac{1}{\sqrt{2}}\). The solutions in the interval \(0 \leq x \leq \pi\) are \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).

(ii) (a) For \(k = 1\), \(f(x) = 3 - 4 \cos x\). The function is increasing from \((0, -1)\) to \((\pi, 7)\). It is not a straight line and flattens at the extremities, indicating an inflection.

(b) The function \(f\) has an inverse because it is one-to-one (1:1) or strictly increasing with no turning points.

(i) The function \(f(x) = 3 - 2 \tan\left(\frac{1}{2}x\right)\) is defined for \(0 \leq x < \pi\). The range of \(\tan\left(\frac{1}{2}x\right)\) is \((-\infty, \infty)\), so \(f(x)\) can take any value less than or equal to 3. Thus, the range is \(f \leq 3\).

(ii) To find \(f\left(\frac{2}{3}\pi\right)\), substitute \(x = \frac{2}{3}\pi\) into the function: \(f\left(\frac{2}{3}\pi\right) = 3 - 2 \tan\left(\frac{1}{2} \times \frac{2}{3}\pi\right) = 3 - 2 \tan\left(\frac{1}{3}\pi\right) = 3 - 2\sqrt{3}\).

(iii) The graph of \(y = f(x)\) starts at \(y = 3\) when \(x = 0\) and decreases as \(x\) increases, with no turning points, tending tangentially towards \(x = \pi\).

(iv) To find \(f^{-1}(x)\), start with \(y = 3 - 2 \tan\left(\frac{x}{2}\right)\). Rearrange to make \(x\) the subject: \(y = 3 - 2 \tan\left(\frac{x}{2}\right) \Rightarrow 2 \tan\left(\frac{x}{2}\right) = 3 - y \Rightarrow \tan\left(\frac{x}{2}\right) = \frac{3-y}{2} \Rightarrow \frac{x}{2} = \tan^{-1}\left(\frac{3-y}{2}\right) \Rightarrow x = 2 \tan^{-1}\left(\frac{3-y}{2}\right)\). Thus, \(f^{-1}(x) = 2 \tan^{-1}\left(\frac{3-x}{2}\right)\).

(i) Solve \(4 - 3 \sin x = 2\) to get \(3 \sin x = 2\), so \(\sin x = \frac{2}{3}\). The solutions are \(x = 0.730\) or \(x = 2.41\).

(ii) The graph of \(y = 4 - 3 \sin x\) should show one complete oscillation, correctly shaped and positioned in the first quadrant.

(iii) The equation \(f(x) = k\) has no solution for \(k < 1\) or \(k > 7\).

(iv) The largest value of \(A\) for which \(g\) has an inverse is \(A = \frac{3\pi}{2}\).

(v) For \(g^{-1}(3)\), solve \(3 = 4 - 3 \sin x\) to get \(\sin x = \frac{1}{3}\). Thus, \(g^{-1}(3) = 2.80\).

(i) The function \(f(x) = 5 - 3 \sin 2x\) is based on the sine function, which oscillates between -1 and 1. Therefore, \(-3 \leq -3 \sin 2x \leq 3\). Adding 5 to each part of the inequality gives \(2 \leq 5 - 3 \sin 2x \leq 8\). Thus, the range of \(f\) is \(2 \leq f(x) \leq 8\).

(ii) The graph of \(y = 5 - 3 \sin 2x\) will show one complete oscillation from \(x = 0\) to \(x = \pi\). The graph starts at \(y = 5\), decreases to \(y = 2\), and then increases back to \(y = 8\) at \(x = \pi\). It is important to note that the graph does not touch the x-axis and is in the first quadrant.

(iii) The function \(f\) does not have an inverse because it is not one-to-one. The function \(f(x) = 5 - 3 \sin 2x\) repeats values within the interval \(0 \leq x \leq \pi\), which means it fails the horizontal line test for invertibility.

(i) The function \(f(x) = 3 - 2 \sin x\) has a range determined by the range of \(\sin x\), which is \([-1, 1]\). Therefore, \(3 - 2(-1) = 5\) and \(3 - 2(1) = 1\). Thus, the range of \(f\) is \(1 \leq f(x) \leq 5\).

(ii) The graph of \(y = 3 - 2 \sin x\) is a sinusoidal wave starting at \(y = 3\), reaching a minimum of \(y = 1\) and a maximum of \(y = 5\), completing one full oscillation from \(x = 0^\circ\) to \(x = 360^\circ\).

(iii) For \(g\) to have an inverse, it must be one-to-one. The largest interval for \(x\) where \(\sin x\) is one-to-one is \(0^\circ \leq x \leq 90^\circ\) or \(0 \leq x \leq \frac{\pi}{2}\).

(iv) To find \(g^{-1}(x)\), set \(y = 3 - 2 \sin x\) and solve for \(x\):

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

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

\(\sin x = \frac{3 - y}{2}\)

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

Thus, \(g^{-1}(x) = \sin^{-1}\left(\frac{3-x}{2}\right)\).

(a) The equation of the curve is \(y = 2 + 3 \, \sin \frac{1}{2}x\). The sine function \(\sin \frac{1}{2}x\) oscillates between -1 and 1. Therefore, the maximum value of \(y\) is when \(\sin \frac{1}{2}x = 1\), giving \(y = 2 + 3 \times 1 = 5\). The minimum value of \(y\) is when \(\sin \frac{1}{2}x = -1\), giving \(y = 2 + 3 \times (-1) = -1\).

(b) The sketch of the curve should show one complete cycle starting and finishing at \(y = 2\), with maximum at \(y = 5\) and minimum at \(y = -1\). The curve should have the correct shape and curvature.

(c) To find the number of solutions for \(2 + 3 \, \sin \frac{1}{2}x = 5 - 2x\), rearrange to \(3 \, \sin \frac{1}{2}x = 3 - 2x\). The left side oscillates between -3 and 3, while the right side is a linear function. There is one intersection point within the given range \(0 \leq x \leq 4\pi\).

(i) The function \(f(x) = 2 - 3 \cos x\) has a range determined by the range of \(\cos x\), which is \([-1, 1]\). Therefore, the range of \(f(x)\) is \(2 - 3(-1) \leq f(x) \leq 2 - 3(1)\), which simplifies to \(-1 \leq f(x) \leq 5\).

(ii) The graph of \(y = f(x)\) is a cosine wave starting at \((0, -1)\), peaking at \((\pi, 5)\), and ending at \((2\pi, -1)\). It is symmetrical about \(x = \pi\).

(iii) For \(g(x) = 2 - 3 \cos x\) to have an inverse, it must be one-to-one. The largest interval for \(x\) where \(\cos x\) is one-to-one is \([0, \pi]\). Thus, the largest value of \(p\) is \(\pi\).

(iv) To find \(g^{-1}(x)\), set \(y = 2 - 3 \cos x\) and solve for \(x\):

\(y = 2 - 3 \cos x\)

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

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

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

(i) Start with the equation \(f(x) + 4 = 0\), which simplifies to \(3 \tan\left(\frac{1}{2}x\right) - 2 + 4 = 0\) or \(3 \tan\left(\frac{1}{2}x\right) = -2\). This gives \(\tan\left(\frac{1}{2}x\right) = -\frac{2}{3}\). Solving for \(\frac{1}{2}x\), we find \(\frac{1}{2}x = \tan^{-1}\left(-\frac{2}{3}\right) \approx -0.588\). Therefore, \(x = 2(-0.588) = -1.2\).

(ii) To find \(f^{-1}(x)\), start with \(y = 3 \tan\left(\frac{1}{2}x\right) - 2\). Rearrange to isolate \(\tan\left(\frac{1}{2}x\right)\): \(y + 2 = 3 \tan\left(\frac{1}{2}x\right)\), so \(\tan\left(\frac{1}{2}x\right) = \frac{y+2}{3}\). Taking the inverse tangent gives \(\frac{1}{2}x = \tan^{-1}\left(\frac{y+2}{3}\right)\), thus \(x = 2 \tan^{-1}\left(\frac{y+2}{3}\right)\). Therefore, \(f^{-1}(x) = 2 \tan^{-1}\left(\frac{x+2}{3}\right)\). The domain of \(f^{-1}\) is \(-5, 1\) based on the range of \(f\).

(iii) The sketch should show a tan graph through the first, third, and fourth quadrants, and an inverse tan graph through the first, second, and third quadrants. The two curves should be symmetrical about \(y = x\) and approximately in the correct domain and range.

(i) The function \(f(x) = 5 - 2 \sin 2x\) has a range determined by the range of \(\sin 2x\), which is \([-1, 1]\). Thus, \(f(x)\) ranges from \(5 - 2(1) = 3\) to \(5 - 2(-1) = 7\). Therefore, the range is \(3 \leq f(x) < 7\).

(ii) The graph of \(y = f(x)\) is a sinusoidal curve with one complete oscillation between \(0\) and \(\pi\), initially going downwards, all above \(f(x) = 0\).

(iii) To solve \(f(x) = 6\), set \(5 - 2 \sin 2x = 6\), giving \(\sin 2x = -\frac{1}{2}\). Solving \(2x = \frac{7\pi}{6}\) or \(2x = \frac{11\pi}{6}\), we find \(x = \frac{7\pi}{12}\) or \(x = \frac{11\pi}{12}\).

(iv) For \(g\) to have an inverse, it must be one-to-one. The largest \(k\) for which this is true is \(\frac{\pi}{4}\).

(v) To find \(g^{-1}(x)\), solve \(5 - 2 \sin 2x = y\) for \(x\). This gives \(\sin 2x = \frac{5-y}{2}\), so \(2x = \sin^{-1} \left( \frac{5-y}{2} \right)\) and \(x = \frac{1}{2} \sin^{-1} \left( \frac{5-x}{2} \right)\).

(i) The function \(f(x) = 4 \sin x - 1\) has a maximum value when \(\sin x = 1\) and a minimum value when \(\sin x = -1\). Therefore, the range of \(f(x)\) is:

\(4(-1) - 1 \leq f(x) \leq 4(1) - 1\)

\(-5 \leq f(x) \leq 3\)

(ii) To find the x-intercept, set \(f(x) = 0\):

\(4 \sin x - 1 = 0\)

\(\sin x = \frac{1}{4}\)

\(x = \sin^{-1}\left(\frac{1}{4}\right) \approx 0.253\)

So, the x-intercept is \((0.253, 0)\).

For the y-intercept, set \(x = 0\):

\(f(0) = 4 \sin 0 - 1 = -1\)

So, the y-intercept is \((0, -1)\).

(iii) The graph of \(y = f(x)\) is a sinusoidal curve starting at \((0, -1)\), peaking at \((\frac{1}{2}\pi, 3)\), and troughing at \((-\frac{1}{2}\pi, -5)\).

(iv) To find the inverse, solve \(y = 4 \sin x - 1\) for \(x\):

\(y + 1 = 4 \sin x\)

\(\sin x = \frac{y + 1}{4}\)

\(x = \sin^{-1}\left(\frac{y + 1}{4}\right)\)

Thus, \(f^{-1}(x) = \sin^{-1}\left(\frac{x + 1}{4}\right)\).

The domain of \(f^{-1}\) is the range of \(f\), \(-5 \leq x \leq 3\), and the range of \(f^{-1}\) is the domain of \(f\), \(-\frac{1}{2}\pi \leq f^{-1}(x) \leq \frac{1}{2}\pi\).

(i) Start with the equation \(5 + 3 \cos\left(\frac{1}{2}x\right) = 7\). Subtract 5 from both sides to get \(3 \cos\left(\frac{1}{2}x\right) = 2\). Divide by 3: \(\cos\left(\frac{1}{2}x\right) = \frac{2}{3}\). Find \(\frac{1}{2}x = \cos^{-1}\left(\frac{2}{3}\right) \approx 0.84\). Multiply by 2 to solve for \(x\): \(x \approx 1.68\).

(ii) The graph of \(y = 5 + 3 \cos\left(\frac{1}{2}x\right)\) is a cosine wave starting at \(y = 8\) when \(x = 0\) and ending at \(y = 2\) when \(x = 2\pi\).

(iii) The function \(f\) has no turning points and is one-to-one over the given interval, so it has an inverse.

(iv) To find \(f^{-1}(x)\), start with \(y = 5 + 3 \cos\left(\frac{1}{2}x\right)\). Rearrange to make \(x\) the subject: \(y - 5 = 3 \cos\left(\frac{1}{2}x\right)\), then \(\cos\left(\frac{1}{2}x\right) = \frac{y-5}{3}\). Solve for \(x\): \(\frac{1}{2}x = \cos^{-1}\left(\frac{y-5}{3}\right)\), so \(x = 2 \cos^{-1}\left(\frac{y-5}{3}\right)\).

(i) Start with the equation \(6 - 4\cos\left(\frac{1}{2}x\right) = 4\). Simplify to \(4\cos\left(\frac{1}{2}x\right) = 2\), giving \(\cos\left(\frac{1}{2}x\right) = \frac{1}{2}\). The solutions for \(\frac{1}{2}x = \frac{1}{3}\pi\) and \(\frac{1}{2}x = \frac{5}{3}\pi\) lead to \(x = \frac{2}{3}\pi\) and \(x = \frac{10}{3}\pi\). However, \(x = \frac{10}{3}\pi\) is outside the given domain, so the valid solutions are \(x = \frac{2}{3}\pi\) and \(x = \frac{4}{3}\pi\).

(ii) The range of \(f(x) = 6 - 4\cos\left(\frac{1}{2}x\right)\) is determined by the range of \(\cos\left(\frac{1}{2}x\right)\), which is \([-1, 1]\). Thus, \(f(x)\) ranges from \(6 - 4(1) = 2\) to \(6 - 4(-1) = 10\), so \(2 \leq f(x) \leq 10\).

(iii) The graph of \(y = f(x)\) is a cosine wave with a point of inflexion at \(x = \pi\).

(iv) To find \(f^{-1}(x)\), start with \(y = 6 - 4\cos\left(\frac{1}{2}x\right)\). Rearrange to \(\cos\left(\frac{1}{2}x\right) = \frac{6-y}{4}\). Then, \(\frac{1}{2}x = \cos^{-1}\left(\frac{6-y}{4}\right)\), leading to \(x = 2\cos^{-1}\left(\frac{6-y}{4}\right)\). Therefore, \(f^{-1}(x) = 2\cos^{-1}\left(\frac{6-x}{4}\right)\).

(i) To solve \(f(x) = 0\), we have:

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

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

Using the inverse cosine function, \(x = \cos^{-1} \left( \frac{2}{3} \right)\), which gives \(x \approx 0.841\) or \(x \approx 5.44\) within the interval \(0 \leq x \leq 2\pi\).

(ii) The range of \(f(x) = 3 \cos x - 2\) is determined by the range of \(\cos x\), which is \([-1, 1]\). Therefore, the range of \(f(x)\) is:

\(3(-1) - 2 \leq f(x) \leq 3(1) - 2\)

\(-5 \leq f(x) \leq 1\)

(iii) The graph of \(y = f(x)\) is a cosine wave starting at \(y = 1\), decreasing to \(y = -5\), and returning to \(y = 1\) over one cycle from \(0\) to \(2\pi\).

(iv) For \(g\) to have an inverse, it must be one-to-one. The maximum interval for \(x\) where \(\cos x\) is one-to-one is \([0, \pi]\). Therefore, \(k = \pi\) or \(180^\circ\).

(v) To find \(g^{-1}(x)\), solve \(y = 3 \cos x - 2\) for \(x\):

\(y + 2 = 3 \cos x\)

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

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

Thus, \(g^{-1}(x) = \cos^{-1} \left( \frac{x + 2}{3} \right)\).