Question

найдите корни уравнения Sin (3x - п/6) = 1/2преднадлежащие промежутку [ -2п; п]

Answer 1

$\sin(3x-\frac{\pi}{6})=\frac{1}{2};\\ 3x-\frac\pi6=(-1)^n\arcsin\frac12+\pi n. n\in Z\\ 3x-\frac\pi6=(-1)^n\frac{\pi}{6}+\pi n, n\in Z\\ 3x=\frac\pi6(1+(-1)^n)+\pi n, n\in Z\\ x=\frac{\pi}{18}(1+(-1)^n)+\frac{\pi n}{3}, n\in Z\\ \left[ {{x=\frac{\pi}{9}+\frac{\pi n}{3},\ n=2k==>x=\frac\pi9+\frac{2\pi k}{3}} \atop {x=\frac{\pi n}{3}, \ n=2k+1==>\ x=\frac\pi3+\frac{2\pi k}{3}}} \right. x\in[-2\pi;\pi]\\ k=-3:x=\frac{\pi}{9}-2\pi=-\frac{17\pi}{9};\ \ x=\frac{\pi}{3}-2\pi=-\frac{5\pi}{3};$
$k=-2:x=\frac\pi9-\frac{4\pi}{3}=-\frac{11\pi}{9};\ \ x=\frac\pi3-\frac{4\pi}{3}=-\pi;\\ k=-1:x=\frac{\pi}{9}-\frac{2\pi}{3}=-\frac{5\pi}{9};\ \ x=\frac{\pi}{3}-\frac{2\pi}{3}-\frac\pi3;\\ k=0:x=\frac\pi9;\ \ \ \frac\pi3;\\ k=1:x=\frac\pi9+\frac{2\pi}{3}=\frac{7\pi}{9};\ \ x=\frac{\pi}{3}+\frac{2\pi}{3}=\pi$

значит имеем такие решения
$x=-\frac{17\pi}{9};\ -\frac{5\pi}{3};\ -\frac{11\pi}{9};\ -\pi;\ -\frac{5\pi}{9};\ -\frac\pi3;\ \frac\pi9;\ \frac\pi3;\ \frac{7\pi}{9};\ \pi$