Question

Найдите все корни уравнения cos(3x-pi/4)=1/2,удовлетворяющие равенству -pi/6 < x < 5pi/6

Answer 1

$cos(3x-\frac{\pi}{4} )=\frac{1}{2}\\\\3x-\frac{\pi}{4}=\pm \frac{\pi}{3}+2\pi n\; ,\; n\in Z\\\\3x=\frac{\pi}{4}\pm \frac{\pi}{3}+2\pi n=\left [ {{\frac{7\pi }{12}+2\pi n\; ,n\in Z} \atop {-\frac{\pi}{12}+2\pi n,\; n\in Z}}\right. \\\\\\x=\frac{\pi }{12}\pm \frac{\pi}{9}+\frac{2\pi n}{3}=\left [ {{\frac{7\pi }{36}+\frac{2\pi n}{3}\, ,\; n\in Z} \atop {-\frac{\pi}{36}+\frac{2\pi n}{3}\, ,\; n\in Z }} \right.\\\\a)\; \; -\frac{\pi}{6}<\frac{7\pi }{36}+\frac{2\pi n}{3}<\frac{5\pi }{6}$

$\frac{-13\pi}{36}<\frac{2\pi n}{3}<\frac{23\pi }{36}\; \; ,\; \; -\frac{13}{24}<n<\frac{23}{24}\; \; \to \; n=0\\\\x=\frac{7\pi }{36}\\\\b)\; \; -\frac{\pi }{6}<-\frac{\pi }{36}+\frac{2\pi n}{3}<\frac{5\pi }{6}\; \; ,\; \; -\frac{5\pi}{36}<\frac{2\pi n}{3}<\frac{31\pi}{36} \; ,\\\\-\frac{5}{24}<n<\frac{31}{24}\; \; ,\; \; -\frac{5}{24}<n<1\frac{7}{24}\; \; \; \to \; \; \; n=0,\; 1\; \\\\x=-\frac{\pi}{36}\; ,\; \; x=-\frac{\pi}{36}+\frac{2\pi }{3}=\frac{23\pi }{36}\\\\Otvet:\; \; \frac{7\pi}{36}\; ,\; \; -\frac{\pi }{36}\; ,\; \; \frac{23\pi }{36}\; .$