First, evaluate \(-\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right)\). The value of \(\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right)\) is \(\frac{\pi}{6}\), so \(-\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right) = -\frac{\pi}{6}\).

Substitute this into the equation:

\(\frac{1}{6}\pi + \arctan(4x) = -\frac{\pi}{6}\).

Rearrange to find \(\arctan(4x)\):

\(\arctan(4x) = -\frac{\pi}{6} - \frac{1}{6}\pi = -\frac{\pi}{3}\).

Thus, \(4x = \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\).

Solving for \(x\), we get:

\(x = -\frac{\sqrt{3}}{4}\).

(i) To find the \(x\)-coordinate of \(A\), set \(\sin x = 2 \cos x\).

This implies \(\tan x = 2\).

Solving for \(x\), we find \(x = 1.11\) (or approximately \(0.352\pi\)).

(ii) For the \(y\)-coordinate of \(B\), we need the value of \(y\) when \(x\) is such that \(y = \sin x = 2 \cos x\) and \(y < 0\).

The negative solution for \(y\) is in the range \(-1 < y < -0.8\).

The value is approximately \(y = -0.895\).

(i) We have the equations:

\(a + \frac{1}{2}b = 5\)

\(a - b = 11\)

Solving these simultaneously:

From \(a - b = 11\), we get \(a = 11 + b\).

Substitute into \(a + \frac{1}{2}b = 5\):

\(11 + b + \frac{1}{2}b = 5\)

\(11 + \frac{3}{2}b = 5\)

\(\frac{3}{2}b = -6\)

\(b = -4\)

Substitute \(b = -4\) into \(a = 11 + b\):

\(a = 11 - 4 = 7\)

Thus, \(a = 7\) and \(b = -4\).

(ii) The function \(f(x) = a + b \cos x\) has a range determined by the maximum and minimum values of \(\cos x\), which are 1 and -1, respectively.

Thus, the range of \(f(x)\) is:

\(a + b \leq f(x) \leq a - b\)

Substitute \(a = 7\) and \(b = -4\):

\(7 - 4 \leq f(x) \leq 7 + 4\)

\(3 \leq f(x) \leq 11\)

For \(f(x) = k\) to have no solution, \(k\) must be outside this range:

\(k < 3\) or \(k > 11\).

The maximum value of the sine function is 1, so at \(\theta = 0^\circ\), \(y = k \sin(\alpha) = 2\). Therefore, \(k \sin(\alpha) = 2\).

At \(\theta = 150^\circ\), the graph crosses the \(\theta\)-axis, so \(y = k \sin(150^\circ + \alpha) = 0\). This implies \(150^\circ + \alpha = 180^\circ\), so \(\alpha = 30^\circ\).

Substituting \(\alpha = 30^\circ\) into \(k \sin(\alpha) = 2\), we have \(k \sin(30^\circ) = 2\).

Since \(\sin(30^\circ) = \frac{1}{2}\), we get \(k \cdot \frac{1}{2} = 2\).

Therefore, \(k = 4\).

(i) The line cuts the x-axis at A, so the y-coordinate of A is 0. The angle \(\angle BAO = \frac{1}{6} \pi\) radians implies that the slope of the line is \(\tan\left(\frac{1}{6} \pi\right) = \frac{1}{\sqrt{3}}\).

The equation of the line is \(y - 2 = \frac{1}{\sqrt{3}}(x - 0)\).

Setting \(y = 0\) to find the x-coordinate of A:

\(0 - 2 = \frac{1}{\sqrt{3}}x\)

\(x = 2\sqrt{3}\)

(ii) The midpoint of AB is \(\left(\frac{\sqrt{3}}{2}, 1\right)\).

The gradient of AB is \(-\frac{1}{\sqrt{3}}\), so the gradient of the perpendicular bisector is \(\sqrt{3}\).

The equation of the perpendicular bisector is \(y - 1 = \sqrt{3}(x - \frac{\sqrt{3}}{2})\).

Simplifying gives \(y = \sqrt{3}x - 2\).

(a)(i) We have the equations:

\(4 = a + \frac{1}{2}b\)

\(3 = a + b\)

Subtract the first equation from the second:

\(3 - 4 = a + b - (a + \frac{1}{2}b)\)

\(-1 = \frac{1}{2}b\)

\(b = -2\)

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

\(3 = a - 2\)

\(a = 5\)

(a)(ii) \(ff(x) = a + b \sin(a + b \sin x)\)

\(ff(0) = 5 - 2 \sin(5) = 6.92\)

(b) The range of \(g(x) = c + d \sin x\) is \(-4 \leq g(x) \leq 10\).

\(10 = c + d\) and \(-4 = c - d\)

Adding these equations:

\(10 - (-4) = c + d + c - d\)

\(14 = 2c\)

\(c = 7\)

Substitute \(c = 3\) into \(10 = c + d\):

\(10 = 3 + d\)

\(d = 7\)

Alternatively, \(d = -7\) also satisfies the range condition.

(i) The graph of \(y = 2 \cos x\) is a cosine wave with amplitude 2. It completes one full cycle from \(-\pi\) to \(\pi\). The point of intersection with the \(y\)-axis is at \(x = 0\), giving \(y = 2 \cos(0) = 2\). Thus, the coordinates are \((0, 2)\).

(ii) The coordinates of \(P\) are \(\left( \frac{\pi}{3}, 2 \cos \left( \frac{\pi}{3} \right) \right) = \left( \frac{\pi}{3}, 1 \right)\) and \(Q\) are \((\pi, 2 \cos(\pi)) = (\pi, -2)\). The distance \(PQ\) is given by:

\(PQ = \sqrt{\left( \pi - \frac{\pi}{3} \right)^2 + (1 - (-2))^2} = \sqrt{\left( \frac{2\pi}{3} \right)^2 + 3^2} = \sqrt{\frac{4\pi^2}{9} + 9}\)

\(PQ = \sqrt{\frac{4\pi^2}{9} + 9} = \sqrt{\frac{4\pi^2 + 81}{9}} = \sqrt{\frac{4\pi^2 + 81}{9}} \approx 3.7\)

(iii) The equation of the line through \(P\) and \(Q\) is:

\(y - 1 = \frac{-2 - 1}{\pi - \frac{\pi}{3}} (x - \frac{\pi}{3})\)

\(y - 1 = \frac{-3}{\frac{2\pi}{3}} (x - \frac{\pi}{3})\)

\(y - 1 = -\frac{9}{2\pi} (x - \frac{\pi}{3})\)

Setting \(y = 0\) to find \(h\):

\(0 - 1 = -\frac{9}{2\pi} (h - \frac{\pi}{3})\)

\(-1 = -\frac{9}{2\pi} h + \frac{3}{2}\)

\(-\frac{5}{2} = -\frac{9}{2\pi} h\)

\(h = \frac{5}{9} \pi\)

Setting \(x = 0\) to find \(k\):

\(k - 1 = -\frac{9}{2\pi} (0 - \frac{\pi}{3})\)

\(k - 1 = \frac{3}{2}\)

\(k = \frac{5}{2}\)

(i) To find the \(x\)-coordinate of \(A\), set \(\tan x = \cos x\).

This implies \(\sin x = \cos^2 x\).

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we have \(\sin x = 1 - \sin^2 x\).

Rearrange to form a quadratic equation: \(\sin^2 x + \sin x - 1 = 0\).

Solving this quadratic equation gives \(\sin x = 0.618\).

Thus, \(x = \sin^{-1}(0.618) = 0.666\).

(ii) For point \(B\), the \(x\)-coordinate is \(\pi - 0.666 = 2.475\).

Calculate the \(y\)-coordinate using \(y = \tan(2.475)\) or \(y = \cos(2.475)\).

This gives \(y = -0.787\).

Therefore, the coordinates of \(B\) are \((2.48, -0.787)\).

(a) Given \(\sin^{-1}(3x) = -\frac{1}{3}\pi\), we have \(3x = \sin\left(-\frac{1}{3}\pi\right) = -\frac{\sqrt{3}}{2}\).

Solving for \(x\), we get \(x = \frac{-\sqrt{3}}{2 \times 3} = -\frac{\sqrt{3}}{6}\).

(b) The equation is \(2 \cos \theta \sin \theta - 2 \cos \theta - \sin \theta + 1 = 0\).

Factorise as \((2\cos \theta - 1)(\sin \theta - 1) = 0\).

This gives \(2\cos \theta - 1 = 0\) or \(\sin \theta - 1 = 0\).

For \(2\cos \theta - 1 = 0\), \(\cos \theta = \frac{1}{2}\), so \(\theta = \frac{\pi}{3}\).

For \(\sin \theta - 1 = 0\), \(\sin \theta = 1\), so \(\theta = \frac{\pi}{2}\).

To find \(c\), we use the fact that the graph crosses the \(x\)-axis at \(C(\cos^{-1} c, 0)\). At this point, \(y = 0\), so:

\(a \, \cos(\cos^{-1} c) - b = 0\)

\(a \, c - b = 0\)

\(c = \frac{b}{a}\)

To find \(d\), we use the fact that the graph crosses the \(y\)-axis at \(D(0, d)\). At this point, \(x = 0\), so:

\(y = a \, \cos(0) - b\)

\(y = a \, (1) - b\)

\(d = a - b\)

Given the equation \(\sin^{-1}(4x^4 + x^2) = \frac{1}{6}\pi\), we first take the sine of both sides to get:

\(4x^4 + x^2 = \sin\left(\frac{1}{6}\pi\right)\).

Since \(\sin\left(\frac{1}{6}\pi\right) = \frac{1}{2}\), we have:

\(4x^4 + x^2 = \frac{1}{2}\).

Let \(y = x^2\), then the equation becomes:

\(4y^2 + y - \frac{1}{2} = 0\).

This is a quadratic equation in \(y\). Solving for \(y\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 1\), \(c = -\frac{1}{2}\).

\(y = \frac{-1 \pm \sqrt{1 + 8}}{8}\).

\(y = \frac{-1 \pm 3}{8}\).

This gives \(y = \frac{1}{4}\) or \(y = -\frac{1}{2}\).

Since \(y = x^2\) and \(y\) must be non-negative, we have \(x^2 = \frac{1}{4}\).

Thus, \(x = \pm \frac{1}{2}\).

Given \(\alpha = \cos^{-1}\left(\frac{8}{17}\right)\), we have \(\cos \alpha = \frac{8}{17}\).

Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we find \(\sin \alpha = \sqrt{1 - \left(\frac{8}{17}\right)^2}\).

Calculate \(\sin \alpha = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}\).

Now, \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{15/17}{8/17} = \frac{15}{8}\).

We need to find \(\frac{1}{\sin \alpha} + \frac{1}{\tan \alpha}\).

\(\frac{1}{\sin \alpha} = \frac{17}{15}\) and \(\frac{1}{\tan \alpha} = \frac{8}{15}\).

Adding these, \(\frac{1}{\sin \alpha} + \frac{1}{\tan \alpha} = \frac{17}{15} + \frac{8}{15} = \frac{25}{15} = \frac{5}{3}\).

(i) The maximum height occurs when \(\cos kt = -1\). Substituting this into the formula gives:

\(h = 60(1 - (-1)) = 60 \times 2 = 120 \text{ m}\)

(ii) Given that one complete revolution takes 30 minutes, \(kt = 2\pi\) when \(t = 30\). Therefore,

\(30k = 2\pi\)

\(k = \frac{2\pi}{30} = \frac{\pi}{15}\)

(iii) To find the time when the height is above 90 m, set \(h = 90\):

\(90 = 60(1 - \cos kt)\)

\(1 - \cos kt = \frac{90}{60} = 1.5\)

\(\cos kt = -0.5\)

The solutions for \(\cos kt = -0.5\) are \(kt = \frac{2\pi}{3}\) or \(kt = \frac{4\pi}{3}\).

Solving for \(t\):

\(t = \frac{2\pi}{3k} = \frac{2\pi}{3 \times \frac{\pi}{15}} = 10 \text{ minutes}\)

\(t = \frac{4\pi}{3k} = \frac{4\pi}{3 \times \frac{\pi}{15}} = 20 \text{ minutes}\)

Thus, the passenger is above 90 m for 10 minutes.

Since \(\theta\) is an obtuse angle, \(\theta\) is in the second quadrant where sine is positive and cosine is negative.

(i) Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\(\cos^2 \theta = 1 - \sin^2 \theta = 1 - k^2\)

\(\cos \theta = -\sqrt{1-k^2}\) because cosine is negative in the second quadrant.

(ii) \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{k}{-\sqrt{1-k^2}} = -\frac{k}{\sqrt{1-k^2}}\)

(iii) \(\sin(\theta + \pi) = -\sin \theta = -k\) because adding \(\pi\) to an angle reflects it across the origin, changing the sign of sine.

First, calculate \(\arctan(3)\). This gives \(\arctan(3) = 1.249\) radians.

Next, we need to find \(\sin(1.249)\). Using a calculator, \(\sin(1.249) \approx 0.949\).

Set \(x - 1 = 0.949\), so \(x = 0.949 + 1 = 1.949\).

Therefore, the value of \(x\) is approximately 1.95.

(i) (a) Since \(\theta\) is a reflex angle, it is in the 4th quadrant where sine is negative. Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - k^2\)

\(\sin \theta = -\sqrt{1-k^2}\)

(i) (b) \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{1-k^2}}{k}\)

(ii) Since \(\theta\) is in the 4th quadrant, \(2\theta\) will be in the range of 540° to 720°, which corresponds to the 3rd and 4th quadrants. In both these quadrants, \(\sin 2\theta\) is negative.

(a) We start with the equation \(\sin^{-1}(x^2 - 1) = \frac{1}{3}\pi\).

This implies \(x^2 - 1 = \sin\left(\frac{\pi}{3}\right)\).

Since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), we have \(x^2 - 1 = \frac{\sqrt{3}}{2}\).

Solving for \(x^2\), we get \(x^2 = 1 + \frac{\sqrt{3}}{2}\).

Thus, \(x = \pm \sqrt{1 + \frac{\sqrt{3}}{2}}\).

Calculating this gives \(x \approx \pm 1.366\) to 3 decimal places.

(b) We have \(\sin(2\theta + \frac{1}{3}\pi) = \frac{1}{2}\).

The general solution for \(\sin A = \frac{1}{2}\) is \(A = \frac{\pi}{6} + 2k\pi\) or \(A = \frac{5\pi}{6} + 2k\pi\), where \(k\) is an integer.

Thus, \(2\theta + \frac{1}{3}\pi = \frac{\pi}{6}\) or \(2\theta + \frac{1}{3}\pi = \frac{5\pi}{6}\).

Solving these equations:

1. \(2\theta + \frac{1}{3}\pi = \frac{\pi}{6}\)

\(2\theta = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}\)

\(\theta = -\frac{\pi}{12}\) (not in the range \(0 \leq \theta \leq \pi\))

2. \(2\theta + \frac{1}{3}\pi = \frac{5\pi}{6}\)

\(2\theta = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\)

\(\theta = \frac{\pi}{4}\)

3. \(2\theta + \frac{1}{3}\pi = \frac{13\pi}{6}\)

\(2\theta = \frac{13\pi}{6} - \frac{\pi}{3} = \frac{13\pi}{6} - \frac{2\pi}{6} = \frac{11\pi}{6}\)

\(\theta = \frac{11\pi}{12}\)

Thus, the solutions are \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{11\pi}{12}\).

(i) Since \(\cos x = p\) and \(x\) is an acute angle, we use the identity \(\sin^2 x + \cos^2 x = 1\). Therefore, \(\sin^2 x = 1 - \cos^2 x = 1 - p^2\). Taking the positive square root (since \(x\) is acute), \(\sin x = \sqrt{1 - p^2}\).

(ii) \(\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{1 - p^2}}{p}\).

(iii) Using the identity \(\tan(90^\circ - x) = \cot x\), we have \(\tan(90^\circ - x) = \frac{1}{\tan x} = \frac{p}{\sqrt{1 - p^2}}\).

(i) We have \(a = \\sin \theta - 3 \\cos \theta\) and \(b = 3 \\sin \theta + \\cos \theta\).

Calculate \(a^2 + b^2\):

\(a^2 = (\\sin \theta - 3 \\cos \theta)^2 = \\sin^2 \theta - 6 \\sin \theta \\cos \theta + 9 \\cos^2 \theta\)

\(b^2 = (3 \\sin \theta + \\cos \theta)^2 = 9 \\sin^2 \theta + 6 \\sin \theta \\cos \theta + \\cos^2 \theta\)

Adding these, \(a^2 + b^2 = \\sin^2 \theta - 6 \\sin \theta \\cos \theta + 9 \\cos^2 \theta + 9 \\sin^2 \theta + 6 \\sin \theta \\cos \theta + \\cos^2 \theta\)

\(= 10 \\sin^2 \theta + 10 \\cos^2 \theta\)

Using \(\\sin^2 \theta + \\cos^2 \theta = 1\), we get \(a^2 + b^2 = 10\).

(ii) Given \(2a = b\), substitute \(a\) and \(b\):

\(2(\\sin \theta - 3 \\cos \theta) = 3 \\sin \theta + \\cos \theta\)

\(2 \\sin \theta - 6 \\cos \theta = 3 \\sin \theta + \\cos \theta\)

\(-\\sin \theta = 7 \\cos \theta\)

\(\\tan \theta = -\frac{1}{7}\)

\(\theta = \\arctan(-\frac{1}{7}) \approx 98.1^\circ\) and \(\theta = 98.1^\circ + 180^\circ = 278.1^\circ\).

(i) We have \(gf(x) = 1\), which means \(g(f(x)) = 1\). Therefore, \(\cos\left(\frac{1}{2}x + \frac{\pi}{6}\right) = 1\).

The cosine function equals 1 at \(0\), so \(\frac{1}{2}x + \frac{\pi}{6} = 0\).

Solving for \(x\), we get:

\(\frac{1}{2}x = -\frac{\pi}{6}\)

\(x = -\frac{\pi}{3}\).

(ii) We have \(fg(x) = 1\), which means \(f(g(x)) = 1\). Therefore, \(\frac{1}{2}\cos x + \frac{\pi}{6} = 1\).

Solving for \(\cos x\), we get:

\(\frac{1}{2}\cos x = 1 - \frac{\pi}{6}\)

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

\(\cos x = 0.9528\)

Solving for \(x\), we find \(x = \pm 0.31\) (to 2 decimal places).

(i) Start with the left-hand side: \(\tan^2 \theta - \sin^2 \theta\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

Thus, \(\tan^2 \theta - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta\).

Factor out \(\sin^2 \theta\):

\(\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right)\).

Using \(1 - \cos^2 \theta = \sin^2 \theta\), we have:

\(\sin^2 \theta \left( \frac{1 - \cos^2 \theta}{\cos^2 \theta} \right) = \sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \tan^2 \theta \sin^2 \theta\).

Thus, \(\tan^2 \theta - \sin^2 \theta \equiv \tan^2 \theta \sin^2 \theta\).

(ii) From the identity, \(\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta\), the right-hand side is positive for \(0^\circ < \theta < 90^\circ\) because \(\tan^2 \theta\) and \(\sin^2 \theta\) are positive.

Therefore, \(\tan^2 \theta > \sin^2 \theta\), implying \(\tan \theta > \sin \theta\) for \(0^\circ < \theta < 90^\circ\).

(i) Given \(f(0) = 10\), we have:

\(a + b = 10\)

Given \(f\left( \frac{2}{3}\pi \right) = 1\), we have:

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

Solving these equations:

From \(a + b = 10\), we get \(a = 10 - b\).

Substitute into \(a - \frac{b}{2} = 1\):

\(10 - b - \frac{b}{2} = 1\)

\(10 - \frac{3b}{2} = 1\)

\(\frac{3b}{2} = 9\)

\(b = 6\)

\(a = 10 - 6 = 4\)

Thus, \(a = 4\) and \(b = 6\).

(ii) The range of \(f(x) = a + b \cos x\) is from \(a - b\) to \(a + b\).

\(a - b = 4 - 6 = -2\)

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

Therefore, the range is \(-2\) to \(10\).

(iii) To find \(f\left( \frac{5}{6}\pi \right)\):

\(\cos\left( \frac{5}{6}\pi \right) = -\cos\left( \frac{1}{6}\pi \right) = -\frac{\sqrt{3}}{2}\)

\(f\left( \frac{5}{6}\pi \right) = 4 + 6\left(-\frac{\sqrt{3}}{2}\right)\)

\(= 4 - 3\sqrt{3}\)

(a) Start with the equation \(3 \cos \theta = 8 \tan \theta\).

Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), substitute to get:

\(3 \cos \theta = 8 \frac{\sin \theta}{\cos \theta}\)

Multiply through by \(\cos \theta\) to eliminate the fraction:

\(3 \cos^2 \theta = 8 \sin \theta\)

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(3(1 - \sin^2 \theta) = 8 \sin \theta\)

Expand and rearrange to form a quadratic equation:

\(3 - 3 \sin^2 \theta = 8 \sin \theta\)

\(3 \sin^2 \theta + 8 \sin \theta - 3 = 0\)

(b) Factor the quadratic equation:

\((3 \sin \theta - 1)(\sin \theta + 3) = 0\)

Solving \(3 \sin \theta - 1 = 0\) gives:

\(\sin \theta = \frac{1}{3}\)

Find \(\theta\) using \(\sin^{-1}\):

\(\theta = \sin^{-1}\left(\frac{1}{3}\right)\)

\(\theta \approx 19.5^\circ\)

(i) Start with \(f(x) = 2 \sin^2 x - 3 \cos^2 x\).

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute to get:

\(f(x) = 2(1 - \cos^2 x) - 3 \cos^2 x\).

Simplify to \(f(x) = 2 - 2 \cos^2 x - 3 \cos^2 x = 2 - 5 \cos^2 x\).

Thus, \(a = 2\) and \(b = -5\).

(ii) The expression \(f(x) = 2 - 5 \cos^2 x\) has its maximum value when \(\cos^2 x = 0\), giving \(f(x) = 2\).

The minimum value occurs when \(\cos^2 x = 1\), giving \(f(x) = 2 - 5 = -3\).

(iii) Solve \(f(x) + 1 = 0\):

\(2 - 5 \cos^2 x + 1 = 0\) simplifies to \(2 - 5 \cos^2 x = -1\).

Thus, \(\cos^2 x = 0.6\).

Solving for \(x\), we find \(x = \cos^{-1}(\pm \sqrt{0.6})\).

This gives \(x \approx 0.685\) and \(x \approx 2.46\).

(i) Using the identity \(\tan(\pi - x) = -\tan x\), we have:

\(\tan(\pi - x) = -k\).

(ii) Using the identity \(\tan\left(\frac{\pi}{2} - x\right) = \cot x = \frac{1}{\tan x}\), we have:

\(\tan\left(\frac{\pi}{2} - x\right) = \frac{1}{k}\).

(iii) Using the identity \(\sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}}\), we have:

\(\sin x = \frac{k}{\sqrt{1 + k^2}}\).

Given \(x = \sin^{-1}\left(\frac{2}{5}\right)\), we have \(\sin x = \frac{2}{5}\).

(i) To find \(\cos^2 x\), use the identity \(\cos^2 x = 1 - \sin^2 x\).

\(\sin^2 x = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\).

Thus, \(\cos^2 x = 1 - \frac{4}{25} = \frac{21}{25}\).

(ii) To find \(\tan^2 x\), use the identity \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\).

\(\tan^2 x = \frac{\frac{4}{25}}{\frac{21}{25}} = \frac{4}{21}\).

(i) We start with the expression \(f(x) = 5 \sin^2 x + 3 \cos^2 x\). Using the identity \(\sin^2 x + \cos^2 x = 1\), we can rewrite \(\cos^2 x\) as \(1 - \sin^2 x\). Thus,

\(f(x) = 5 \sin^2 x + 3(1 - \sin^2 x) = 5 \sin^2 x + 3 - 3 \sin^2 x = 3 + 2 \sin^2 x.\)

Therefore, \(a = 3\) and \(b = 2\).

(ii) We need to solve \(3 + 2 \sin^2 x = 7 \sin x\). Let \(s = \sin x\), then the equation becomes:

\(3 + 2s^2 = 7s.\)

Rearranging gives:

\(2s^2 - 7s + 3 = 0.\)

Using the quadratic formula \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = -7, c = 3\), we find:

\(s = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}.\)

This gives \(s = 3\) or \(s = \frac{1}{2}\). Since \(s = \sin x\) and \(0 \leq x \leq \pi\), \(s = 3\) is not possible. Thus, \(s = \frac{1}{2}\), which gives \(x = \frac{\pi}{6}\) or \(x = \frac{5\pi}{6}\).

(iii) The minimum value of \(f(x) = 3 + 2 \sin^2 x\) occurs when \(\sin^2 x = 0\), giving \(f(x) = 3\). The maximum value occurs when \(\sin^2 x = 1\), giving \(f(x) = 5\). Thus, the range of \(f\) is \(3 \leq f(x) \leq 5\).

(a) The equation \(3 \tan^2 x - 5 \tan x - 2 = 0\) can be factored as \((\tan x - 2)(3 \tan x + 1) = 0\). This gives \(\tan x = 2\) or \(\tan x = -\frac{1}{3}\).

For \(\tan x = 2\), \(x = 63.4^\circ\) (only value in range).

For \(\tan x = -\frac{1}{3}\), \(x = 161.6^\circ\) (only value in range).

(b) For the equation \(3 \tan^2 x - 5 \tan x + k = 0\) to have no solutions, the discriminant must be negative: \(b^2 - 4ac < 0\).

Here, \(b = -5\), \(a = 3\), \(c = k\). So, \((-5)^2 - 4(3)(k) < 0\) simplifies to \(25 - 12k < 0\), leading to \(k > \frac{25}{12}\).

(c) For three solutions, the discriminant must be zero, and \(k = 0\).

Substitute \(k = 0\) into the equation: \(3 \tan^2 x - 5 \tan x = 0\).

Factor to get \(\tan x (3 \tan x - 5) = 0\), giving \(\tan x = 0\) or \(\tan x = \frac{5}{3}\).

For \(\tan x = 0\), \(x = 0^\circ\) or \(x = 180^\circ\).

For \(\tan x = \frac{5}{3}\), \(x = 59.0^\circ\).

(i) We start with the function \(f(x) = 3 \cos^2 x - 2 \sin^2 x\).

Using the identity \(\cos^2 x + \sin^2 x = 1\), we can express \(\sin^2 x\) as \(1 - \cos^2 x\).

Substitute \(\sin^2 x = 1 - \cos^2 x\) into the function:

\(f(x) = 3 \cos^2 x - 2(1 - \cos^2 x)\)

\(= 3 \cos^2 x - 2 + 2 \cos^2 x\)

\(= 5 \cos^2 x - 2\)

Thus, \(a = 5\) and \(b = -2\).

(ii) To find the range of \(f\), note that \(\cos^2 x\) ranges from 0 to 1 for \(0 \leq x \leq \pi\).

Substitute the extreme values of \(\cos^2 x\) into \(f(x) = 5 \cos^2 x - 2\):

When \(\cos^2 x = 0\), \(f(x) = 5(0) - 2 = -2\).

When \(\cos^2 x = 1\), \(f(x) = 5(1) - 2 = 3\).

Therefore, the range of \(f\) is \([-2, 3]\).

(i) The function \(f(x) = p \sin^2 2x + q\) has a range determined by the values of \(\sin^2 2x\), which vary from 0 to 1. Therefore, the minimum value of \(f(x)\) is \(q\) and the maximum value is \(p + q\). Thus, the range is \(q \leq f(x) \leq p + q\).

(ii) (a) For \(f(x) = p + q\), \(\sin^2 2x = 1\). This occurs at two points within the interval \(0 \leq x \leq \pi\), so there are 2 solutions.

(b) For \(f(x) = q\), \(\sin^2 2x = 0\). This occurs at three points within the interval \(0 \leq x \leq \pi\), so there are 3 solutions.

(c) For \(f(x) = \frac{1}{2}p + q\), \(\sin^2 2x = \frac{1}{2}\). This occurs at four points within the interval \(0 \leq x \leq \pi\), so there are 4 solutions.

(iii) Given \(p = 3\) and \(q = 2\), solve \(3 \sin^2 2x + 2 = 4\) which simplifies to \(\sin^2 2x = \frac{2}{3}\).

\(\sin 2x = \pm \sqrt{\frac{2}{3}}\). Solving for \(2x\), we find \(2x = \sin^{-1}(\pm \sqrt{\frac{2}{3}})\).

Using a calculator, \(2x \approx 0.955, 2.18, 4.09, 5.32\).

Thus, \(x \approx 0.478, 1.09, 2.05, 2.66\).

(i) For the curve \(y = 3 \cos 2x\), the maximum value of \(\cos 2x\) is 1 and the minimum is -1. Therefore, the largest value of \(y\) is 3 and the smallest is -3.

For the line \(2y + \frac{3x}{\pi} = 5\), rearrange to find \(y = \frac{5}{2} - \frac{3x}{2\pi}\). The smallest value of \(y\) occurs when \(x = 2\pi\), giving \(y = -\frac{1}{2}\), and the largest value occurs when \(x = 0\), giving \(y = \frac{5}{2}\).

(ii) The sketch should show two complete oscillations of the cosine curve starting with a maximum at \((0, 3)\) and leveling off at \(0\) and \(2\pi\). The line should start on the positive \(y\)-axis and finish below the \(x\)-axis at \(2\pi\).

(iii) The number of solutions of the equation \(6 \cos 2x = 5 - \frac{3x}{\pi}\) for \(0 \leq x \leq 2\pi\) is 4.

(i) We know that \(\sin^2 x + \cos^2 x = 1\).

Substitute \(\sin x = a + b\) and \(\cos x = a - b\):

\((a + b)^2 + (a - b)^2 = 1\)

Expanding both terms, we get:

\(a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 1\)

Combine like terms:

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

Divide by 2:

\(a^2 + b^2 = \frac{1}{2}\)

(ii) Given \(\tan x = 2\), we have:

\(\tan x = \frac{\sin x}{\cos x} = \frac{a + b}{a - b} = 2\)

Cross-multiply to solve for a:

\(a + b = 2(a - b)\)

\(a + b = 2a - 2b\)

Rearrange terms:

\(b + 2b = 2a - a\)

\(3b = a\)

Thus, \(a = 3b\).

Given the equation \(y = a + \tan bx\), we know the curve intersects the \(y\)-axis at \((0, \sqrt{3})\). Substituting \(x = 0\) into the equation gives:

\(\sqrt{3} = a + \tan(0)\)

\(\sqrt{3} = a\)

Thus, \(a = \sqrt{3}\).

Next, the curve intersects the \(x\)-axis at \(\left(-\frac{1}{6}\pi, 0\right)\). Substituting \(y = 0\) and \(x = -\frac{1}{6}\pi\) into the equation gives:

\(0 = \sqrt{3} + \tan\left(-\frac{b}{6}\pi\right)\)

\(\tan\left(-\frac{b}{6}\pi\right) = -\sqrt{3}\)

The angle whose tangent is \(-\sqrt{3}\) is \(-\frac{\pi}{3}\), so:

\(-\frac{b}{6}\pi = -\frac{\pi}{3}\)

Solving for \(b\):

\(\frac{b}{6} = \frac{1}{3}\)

\(b = 2\)

Thus, \(b = 2\).