\[ \text{Calculate the maximum and minimum values of the equation} \] \[ y = 4\cos(2x)^\circ + 3\sin(2x)^\circ + 11, \qquad 0 \le x \le 360^\circ. \]

\[ 4\cos(2x)^\circ + 3\sin(2x)^\circ = k\sin(2x + \alpha) \] \[ k = \sqrt{4^2 + 3^2} = 5 \] \[ \therefore\quad -5 \le 4\cos(2x)^\circ + 3\sin(2x)^\circ \le 5 \] \[ \text{The amplitude of the sine wave is } 5, \]

\[ \qquad 4\cos(2x)^\circ + 3\sin(2x)^\circ = 5\sin(2x + \alpha) \] \[ \text{So} \qquad 4\cos(2x)^\circ + 3\sin(2x)^\circ + 11 = 5\sin(2x + \alpha) + 11 \] \[ \text{so the maximum value is } 5 + 11 = 16, \] \[ \text{and the minimum value is } -5 + 11 = 6. \]

\[ \text{Find the maximum value of } 12 + \sqrt{3}\sin x + \cos x \] \[ \text{and determine the corresponding value of } x \] \[ \text{in the range } 0 \le x \le 360^\circ. \]

\[ \text {Let } \sqrt{3}\sin x + \cos x = k\sin(x+\alpha) \] \[ \sqrt{3}\sin x + \cos x = k\bigl(\sin x \cos\alpha + \cos x \sin\alpha\bigr) \] \[ \Rightarrow\; \sqrt{3}\sin x = k\sin x \cos\alpha, \qquad \cos x = k\cos x \sin\alpha \] \[ \Rightarrow\; \sqrt{3} = k\cos\alpha, \qquad 1 = k\sin\alpha \] \[ \text{So } a=\sqrt{3},\; b=1 \] \[ k = \sqrt{a^2 + b^2} \] \[ k = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

\[ \text{Since both } \cos\alpha \text{ and } \sin\alpha \text{ are positive,} \quad \alpha \text{ lies in the first quadrant.} \] \[ \tan\alpha = \frac{b}{a} \] \[ \Rightarrow\; \tan\alpha = \frac{1}{\sqrt{3}} \] \[ \Rightarrow\; \alpha = \tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \] \[ \sqrt{3}\sin x + \cos x = 2\sin(x + 30^\circ) \]

\[ 12 + \sqrt{3}\sin\theta + \cos\theta = 12 + 2\sin(\theta + 30^\circ) \] \[ \text{Max value of } \sin(\theta + 30^\circ) = 1 \] \[ \Rightarrow\; \text{Max value of } 2\sin(\theta + 30^\circ) = 2 \] \[ \Rightarrow\; \text{Max value of } 12 + \sqrt{3}\sin\theta + \cos\theta = 12 + 2 = 14 \] \[ \text{This occurs when } \sin(\theta + 30^\circ) = 1 \] \[ \theta + 30^\circ = 90^\circ \] \[ \Rightarrow\; \theta = 60^\circ \] \[ \boxed{y_{\max} = 14 \text{ at } x = 60^\circ} \]

\[ \text{Sketch the graph of the equation } \sqrt{27}\,\sin x + 3\cos x \text{ in the form } k\cos(x - \alpha), \] \[ \text{where } 0 \le \alpha \le 2\pi. \]

\[ \text{Let } \sqrt{27}\,\sin x + 3\cos x = k\cos(x - \alpha) \] \[ \sqrt{27}\,\sin x + 3\cos x = k\bigl(\cos x\cos\alpha + \sin x\sin\alpha\bigr) \] \[ \Rightarrow\; \sqrt{27}\,\sin x = k\sin\alpha\,\sin x, \qquad 3\cos x = k\cos\alpha\,\cos x \] \[ \Rightarrow\; \sqrt{27} = k\sin\alpha, \qquad 3 = k\cos\alpha \] \[ \text{So } a = 3,\quad b = \sqrt{27} \quad\text{(since } a = k\cos\alpha,\; b = k\sin\alpha\text{)} \] \[ k = \sqrt{a^2 + b^2} \] \[ k = \sqrt{3^2 + (\sqrt{27})^2} = \sqrt{9 + 27} = \sqrt{36} = 6 \]

\[ \text{Since both } \cos\alpha \text{ and } \sin\alpha \text{ are positive, } \alpha \text{ lies in the first quadrant.} \] \[ \tan\alpha = \frac{b}{a} \] \[ \Rightarrow\; \tan\alpha = \frac{\sqrt{27}}{3} \] \[ \Rightarrow\; \alpha = \tan^{-1}\!\left(\frac{\sqrt{27}}{3}\right) \quad\text{(remember to use radians)} \] \[ \Rightarrow\; \alpha = 1.0471975\ \text{rads} \] \[ \Rightarrow\; \alpha = \frac{\pi}{3}\ \text{rads} \] \[ \sqrt{27}\,\sin x + 3\cos x = 6\cos\!\left(x - \frac{\pi}{3}\right) \]

\[ \cos\!\left(x - \frac{\pi}{3}\right) \text{ has a maximum value of } 1 \text{ and a minimum value of } -1. \] \[ \text{So the maximum value of } 6\cos\!\left(x - \frac{\pi}{3}\right) \text{ is } 6 \] \[ \text{and the minimum value is } -6. \] \[ \text{The maximum value occurs when } \cos\!\left(x - \frac{\pi}{3}\right) = 1. \] \[ \Rightarrow\; x - \frac{\pi}{3} = \cos^{-1}(1) \quad\text{(remember to use radians)} \] \[ \Rightarrow\; x - \frac{\pi}{3} = 0,\; 2\pi \] \[ \Rightarrow\; x = 0 + \frac{\pi}{3} \quad\text{and}\quad x = 2\pi + \frac{\pi}{3} \] \[ \Rightarrow\; x = \frac{\pi}{3}, \qquad x = \frac{7\pi}{3}. \]

\[ \text{The minimum value occurs when } \cos\!\left(x - \frac{\pi}{3}\right) = -1 \] \[ \Rightarrow\; x - \frac{\pi}{3} = \cos^{-1}(-1) \quad\text{(remember to use radians)} \] \[ \Rightarrow\; x - \frac{\pi}{3} = \pi \] \[ \Rightarrow\; x = \pi + \frac{\pi}{3} \] \[ \Rightarrow\; x = \frac{4\pi}{3} \]

\[ \text{Maximum value of } 6\cos\!\left(x - \frac{\pi}{3}\right) \text{ is } 6 \] \[ \text{which occurs at } x = \frac{\pi}{3},\quad x = \frac{7\pi}{3}. \] \[ \text{Minimum value of } 6\cos\!\left(x - \frac{\pi}{3}\right) \text{ is } -6 \] \[ \text{which occurs at } x = \frac{4\pi}{3}. \]

\[ \text{Solve } 2\cos x^\circ + 3\sin x^\circ = -1 \qquad 0 \le x \le 360^\circ. \]

\[ \text{Let } 2\cos x^\circ + 3\sin x^\circ = k\cos(x^\circ - \alpha^\circ) \] \[ 2\cos x^\circ + 3\sin x^\circ = k\bigl(\cos x^\circ \cos\alpha^\circ + \sin x^\circ \sin\alpha^\circ\bigr) \] \[ \Rightarrow\; 2\cos x^\circ = k\cos\alpha^\circ \cos x^\circ, \qquad 3\sin x^\circ = k\sin\alpha^\circ \sin x^\circ \] \[ \Rightarrow\; 2 = k\cos\alpha^\circ, \qquad 3 = k\sin\alpha^\circ \] \[ \text{So } a = 2,\quad b = 3 \quad\text{(since } a = k\cos\alpha,\; b = k\sin\alpha\text{)} \] \[ k = \sqrt{a^2 + b^2} \] \[ k = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

\[ \text{Since both } \cos\alpha \text{ and } \sin\alpha \text{ are positive, } \alpha \text{ lies in the first quadrant.} \] \[ \tan\alpha = \frac{b}{a} \] \[ \Rightarrow\; \tan\alpha = \frac{3}{2} \] \[ \Rightarrow\; \alpha = \tan^{-1}\!\left(\frac{3}{2}\right) \] \[ \Rightarrow\; \alpha = 56.31^\circ \; (2\text{ d.p.}) \] \[ 2\cos x^\circ + 3\sin x^\circ = \sqrt{13}\,\cos\!\left(x - 56.31^\circ\right) \]

\[ 2\cos x^\circ + 3\sin x^\circ = -1 \] \[ \Rightarrow\; \sqrt{13}\,\cos\!\left(x - 56.31^\circ\right) = -1 \] \[ \Rightarrow\; \cos\!\left(x - 56.31^\circ\right) = \frac{-1}{\sqrt{13}} \] \[ \Rightarrow\; x - 56.31^\circ = \cos^{-1}\!\left(\frac{-1}{\sqrt{13}}\right) \] \[ \Rightarrow\; x - 56.31^\circ = 106.10^\circ \quad\text{and}\quad 253.90^\circ \] \[ \Rightarrow\; x = 106.10^\circ + 56.31^\circ \quad\text{and}\quad x = 253.90^\circ + 56.31^\circ \] \[ \Rightarrow\; x = 162.41^\circ, \qquad x = 310.21^\circ. \]

\[ \text{Solve } 2\cos 2x^\circ + 3\sin2 x^\circ = -1 \qquad 0 \le x \le 360^\circ. \]

\[ \text{Let } 2\cos 2x^\circ + 3\sin 2x^\circ = k\cos(2x^\circ - \alpha^\circ) \] \[ 2\cos 2x^\circ + 3\sin 2x^\circ = k\bigl(\cos 2x^\circ \cos\alpha^\circ + \sin 2x^\circ \sin\alpha^\circ\bigr) \] \[ \Rightarrow\; 2\cos 2x^\circ = k\cos\alpha^\circ \cos 2x^\circ, \qquad 3\sin 2x^\circ = k\sin\alpha^\circ \sin 2x^\circ \] \[ \Rightarrow\; 2 = k\cos\alpha^\circ, \qquad 3 = k\sin\alpha^\circ \] \[ \text{So } a = 2,\quad b = 3 \quad\text{(since } a = k\cos\alpha,\; b = k\sin\alpha\text{)} \] \[ k = \sqrt{a^2 + b^2} \] \[ k = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

\[ \text{Since both } \cos\alpha \text{ and } \sin\alpha \text{ are positive, } \alpha \text{ lies in the first quadrant.} \] \[ \tan\alpha = \frac{b}{a} \] \[ \Rightarrow\; \tan\alpha = \frac{3}{2} \] \[ \Rightarrow\; \alpha = \tan^{-1}\!\left(\frac{3}{2}\right) \] \[ \Rightarrow\; \alpha = 56.31^\circ \; (2\text{ d.p.}) \] \[ 2\cos 2x^\circ + 3\sin 2x^\circ = \sqrt{13}\,\cos\!\left(2x^\circ - 56.31^\circ\right) \]

\[ 2\cos 2x^\circ + 3\sin 2x^\circ = -1 \] \[ \Rightarrow\; \sqrt{13}\,\cos\!\left(2x^\circ - 56.31^\circ\right) = -1 \] \[ \Rightarrow\; \cos\!\left(2x^\circ - 56.31^\circ\right) = \frac{-1}{\sqrt{13}} \] \[ \Rightarrow\; 2x^\circ - 56.31^\circ = \cos^{-1}\!\left(\frac{-1}{\sqrt{13}}\right) \] \[ \Rightarrow\; 2x^\circ - 56.31^\circ = 106.10^\circ \quad\text{and}\quad 253.90^\circ \]

\[ \text{This gives the first cycle, but remember that} \] \[ \text{there are two cycles in this period.} \] \[ \text{Find the other values by adding } 360^\circ. \] \[ (2x - 56.31^\circ) = 106.10^\circ + 360^\circ = 466.10^\circ, \] \[ (2x - 56.31^\circ) = 253.90^\circ + 360^\circ = 613.90^\circ. \]