Question

Sin 2x + sin 6x - cos 2x = 0 и вот ещё: f(x) = ln (5x-4/2x-1) , f'(1)

Answer 1

$1)\; \; (sin2x+sin6x)-cos2x=0\\\\2sin4x\cdot cos2x-cos2x=0\\\\cos2x\cdot (2sin4x-1)=0\\\\cos2x=0\; ,\; 2x=\frac{\pi}{2}+\pi n,\; x=\frac{\pi}{4}+\frac{\pi n}{2}\; ,\; n\in Z\\\\sin4x=\frac{1}{2}\; ,\; 4x=(-1)^{m}\frac{\pi}{6}+\pi m,\; x=(-1)^{m}\frac{\pi}{24}+\frac{\pi m}{4}\; ,\; m\in Z\\\\2)\; \; f(x)=ln\frac{5x-4}{2x-1}\\\\f'(x)=\frac{2x-1}{5x-4}\cdot \frac{5(2x-1)-2(5x-4)}{(2x-1)^2}= \frac{2x-1}{5x-4} \cdot \frac{3}{(2x-1)^2} = \frac{3}{(5x-4)(2x-1)} \\\\f'(1)=\frac{3}{1\cdot 1}=3$