Question

ctg 4x + 1 = 0

sin 8x - sin 2x = 0

Answer 1

$ctg4x+1=0\\ctg4x=-1\\4x=-\frac{\pi}4+\pi n;n\in Z\\x=-\frac{\pi}{16}+\frac{\pi n}4;n\in Z$

$sin8x-sin2x=0\\\\(sin(a)-sin(b)=2*sin\frac{a-b}2*cos\frac{a+b}2)\\\\2*sin\frac{8x-2x}2*cosx\frac{8x+2x}2=0\\sin\frac{6x}2*cos\frac{10x}2=0\\sin3x*cos5x=0\\\\ \left[\begin{array}{ccc}sin3x=0\\cos5x=0\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}3x=\pi n;n\in Z\\5x=\frac{\pi}2+\pi n;n\in Z\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}x=\frac{\pi n}3;n\in Z\\x=\frac{\pi}{10}+\frac{\pi n}5;n\in Z\end{array}\right$