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Question 1

Question 2

Решите задачу:

$tg (\, 3x) = ctg (\, 4x) \\ \\ \frac{\sin{3x}}{\cos{3x}}=\frac{\cos{4x}}{\sin{4x}} \\ \\ \sin{3x} \cdot \sin{4x}=\cos3x \cdot \cos4x \\ \\ \frac{1}{2} \cdot (\cos{(3x-4x)} - \cos{(3x+4x)} )= \frac{1}{2} \cdot (\cos{(3x-4x)} + \cos{(3x+4x)} ) \\ \\ \frac{1}{2} \cdot \cos{(-x)} - \frac{1}{2} \cdot \cos{7x} = \frac{1}{2} \cdot \cos{(-x)} + \frac{1}{2} \cdot \cos{7x} \\ \\ \frac{1}{2} \cdot \cos{x} - \frac{1}{2} \cdot \cos{7x} = \frac{1}{2} \cdot \cos{x} + \frac{1}{2} \cdot \cos{7x}$

$\\ \\ - \frac{1}{2} \cdot \cos{7x} = \frac{1}{2} \cdot \cos{7x} \\ \\ - \frac{1}{2} \cdot \cos{7x} - \frac{1}{2} \cdot \cos{7x} =0 \\ \\ - \cos{7x}=0 \\ \\ \cos{7x}=0 \\ \\ 7x=\frac{\pi}{2}+\pi n , \ n \in Z \\ \\ x=\frac{\pi}{14}+ \frac{\pi n}{7} , \ n \in Z$

Question 3

$\textup{tg3x=ctg4x} \\ \\ \textup{tg3x= } \frac{1}{\textup{tg4x}} \\ \\ \textup{tg3x}\cdot\textup{tg4x}=1 \\ \\ \frac{\textup{cos(3x-4x)-cos(3x+4x)}}{\textup{cos(3x-4x)+cos(3x+4x)}} =1 \\ \\ \frac{\textup{cosx-cos7x}}{\textup{cosx+cos7x}} =1 \\ \\ \frac{-2sin4x*sin(-3x)}{2cos4x*cos3x} =1 \\ \\ \frac{2sin4x*sin3x}{2cos4x*cos3x} =1$

$\textup{2sin4x sin3x = 2cos4x cos3x} \\ \textup{sin4x sin3x = cos4x cos3x} \\ \textup{sin4x sin3x}-\textup{cos4x cos3x}=0 \\ -(\textup{cos4x cos3x}-\textup{sin4x sin3x})=0 \\ -\textup{cos(4x+3x)}=0 \\ \textup{cos(4x+3x)}=0 \\ \textup{cos7x}=0 \\ \textup{7x}= \frac{ \pi }{\textup{2}}+ \pi \textup{k} \\ \textup{x}= \frac{ \pi }{\textup{14}} + \frac{ \pi \textup{k}}{7},~~k\in Z.$

Ответ: $\frac{ \pi }{\textup{14}} + \frac{ \pi \textup{k}}{7},~~k\in Z$