Problems

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*}