Problems

Solution: \begin{align*} && \cos(A \pm B) &= \cos A \cos B \mp \sin A \sin B \\ A = a, B = 3a&& \cos 4a + \cos 2a &= 2\cos 3a \cos a \\ \Rightarrow && \cos a \cos 3a &= \tfrac12(\cos 4a + \cos 2a) \\ \\ && \sin(A \pm B) &= \sin A \cos B \pm \cos A \sin B \\ && \sin 4a + \sin(- 2a) &= 2 \sin a \cos 3a \\ \Rightarrow && \sin a \cos 3a &= \tfrac12 (\sin 4a - \sin 2a) \end{align*}

1. \(\,\) \begin{align*} && 1 &= 4 \cos x \cos 2x \cos 3x \\ &&&= 2(\cos 4x +\cos 2x)\cos 2x \\ c = \cos 2x:&&&= 2(2c^2-1+c)c \\ \Rightarrow && 0 &= 4c^3+2c^2-2c-1 \\ &&&= (2c+1)(2c^2-1) \\ \Rightarrow && \cos 2x &= -\frac12 \\ \Rightarrow && x &= \frac{\pi}{3}, \frac{2\pi}{3} \\ && \cos 2x &= \pm \frac1{\sqrt2} \\ \Rightarrow && x&= \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \end{align*}
2. \(\,\) \begin{align*} && \tan x &= \tan 2x \tan 3x \tan 4x \\ \Rightarrow &&1 &= \frac{\cos x\sin 2x \sin 3x \sin 4x}{\sin x \cos 2x \cos 3x \cos 4x} \\ &&&= \frac{\sin 2x \sin 4x (\sin4 x + \sin 2x)}{\cos 2x \cos 4x (\sin 4x - \sin 2x)} \\ &&&= \frac{(\cos 2x - \cos 6x) (\sin4 x + \sin 2x)}{(\cos 6x + \cos 2x) (\sin 4x - \sin 2x)} \\ \Rightarrow && 0 &= 2\cos 6x \sin 4x - 2\cos 2x \sin 2 x\\ &&&= \sin 4 x (2 \cos 6x - 1) \\ \Rightarrow && \sin 4x &= 0 \\ \text{ or }&& \cos 6x &= \frac12 \end{align*} \(\sin 4x = 0 \Rightarrow x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\) \(\cos 6x = \frac12 \Rightarrow x = \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\). We should verify these work, since not all of them will, especially where \(\sin 4x = 0\), so our final answer is \(x = 0, \pi, \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\)

Solution: \begin{align*} && I &= \int_0^{2 \pi} e^{x} \cos m x \d x \\ &&&= \left [e^x \cos m x \right]_0^{2 \pi}-\int_0^{2 \pi} e^x m (-\sin mx) \d x\\ &&&= e^{2\pi}-1 + m\int_0^{2\pi}e^x \sin m x \d x \\ &&&= e^{2\pi}-1 + m\left [e^x \sin m x \right]_0^{2\pi} - m \int_0^{2\pi} e^x m \cos x \d x \\ &&&= e^{2\pi}-1+0 - m^2 I\\ \Rightarrow && (m^2+1)I &= e^{2\pi}-1 \\ \Rightarrow && I &= \frac{1}{m^2+1} (e^{2\pi}-1) \end{align*}

1. \(\,\) \begin{align*} && \cos(A+B) + \cos(A-B) &= \cos A\cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B \\ &&&= 2 \cos A \cos B \end{align*} Therefore \begin{align*} && I &= \int_0^{2\pi} e^x \cos x \cos 6x \d x \\ &&&= \int_0^{2\pi} e^x \frac12\left (\cos 7x + \cos 5x \right) \d x\\ &&&= \left ( \frac{1}{2(1+7^2)} + \frac1{2(1+5^2)}\right)(e^{2\pi}-1) \\ &&&= \left (\frac{1}{100}+\frac{1}{52} \right) (e^{2\pi}-1) \\ &&&= \frac{19}{650}(e^{2\pi}-1) \end{align*}
2. \(\,\) \begin{align*} && I &= \int_0^{2\pi} e^x \sin 2x \sin 4x \cos x \d x\\ &&&= \int_0^{2\pi} e^x \tfrac12(\cos2x-\cos 6x) \cos x \d x\\ &&&= \frac12 \int_0^{2\pi} e^x \left (\cos 2x \cos x -\cos 6x \cos x \right) \d x \\ &&&= \frac14 \int_0^{2\pi} e^x \left (\cos 3x + \cos x-\cos 7x-\cos 5x \right) \d x \\ &&&= \frac14 \left (\frac{1}{1+3^2}+\frac{1}{1+1^2}-\frac{1}{1+7^2} - \frac{1}{1+5^2} \right)(e^{2\pi}-1) \\ &&&= \frac14 \left (\frac{1}{10}+\frac{1}{2}-\frac{1}{50} - \frac{1}{26} \right)(e^{2\pi}-1) \\ &&&= \frac{44}{325}(e^{2\pi}-1) \end{align*}

Solution: \begin{align*} && \alpha \sin(A-B) + \beta \cos (A + B) &= \gamma \sin(A+B) \\ \Leftrightarrow && \alpha \sin A \cos B - \alpha \cos A \sin B + \beta \cos A \cos B - \beta \sin A \sin B &= \gamma \sin A \cos B + \gamma \cos A \sin B \\ \Leftrightarrow && \alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B &= \gamma \tan A + \gamma \tan B \\ \Leftrightarrow && \beta \tan A \tan B +(\gamma-\alpha) \tan A + (\gamma +\alpha)\tan B&=\beta \\ \Leftrightarrow && \tan A \tan B +\left (\frac{\gamma-\alpha}{\beta} \right) \tan A + \left (\frac{\gamma +\alpha}{\beta} \right)\tan B&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) - \frac{\gamma^2 - \alpha^2}{\beta^2}&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) &= \frac{\beta^2+\gamma^2-\alpha^2}{\beta^2}\\ \end{align*} Which has the desired form iff \(\beta^2+\gamma^2 = \alpha^2\).

1. \(\,\) \begin{align*} && 2\sin(x-\tfrac14\pi) + \sqrt 3 \cos(x+\tfrac14\pi) &=\sin(x+\tfrac14\pi) \\ && 3 + 1 &= 4 \\ \Rightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{4} + \frac{1-2}{\sqrt3}\right) &= 0\\ \Rightarrow && \tan x &= -\sqrt3 \\ \Rightarrow && x &= \tfrac23\pi, \tfrac53\pi \end{align*}
2. \(\,\) \begin{align*} && 2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) &=\sin(x+\frac16\pi) \\ \Leftrightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{3} + \frac{1-2}{\sqrt3}\right) &= 0\\ && x &\in [0, 2\pi) \end{align*}
3. \(\,\) \begin{align*} && 2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x) \\ && A-B =x + \tfrac13\pi, A+B &= 3x \\ \Rightarrow && A = 2x + \tfrac\pi6, B &= x-\tfrac{\pi}{6} \\ \Rightarrow && \tan (2x+\tfrac\pi6)&=-\sqrt3 \\ && 2x + \tfrac{\pi}{6} &= \tfrac23\pi, \tfrac53\pi, \tfrac83 \pi, \tfrac{11}3\pi \\ && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \\ && \tan(-x-\tfrac{\pi}{6}) &= \frac1{\sqrt{3}} \\ \Rightarrow && x-\tfrac{\pi}{6} &= \ldots, \tfrac{\pi}{6}, \tfrac{7\pi}{6}, \ldots \\ \Rightarrow && x &= \tfrac{\pi}3, \tfrac{4\pi}{3} \\ \\ \Rightarrow && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} , \tfrac{\pi}3, \tfrac{4\pi}{3} \end{align*}