1)Решите уравнение: sin2x=cos(3п/2+x) 2) Найдите все корни, принадлежащие промежутку...

1)
$\sin2x=\cos(\frac{3\pi}{2}+x);\\ 2\sin x\cos x=\cos\frac{3\pi}{2}\cdot\cos x-\sin\frac{3\pi}{2}\cdot\sin x;\\ |\sin\frac{3\pi}{2}=-1;\ \ \ \ \cos\frac{3\pi}{2}=0;|\\ 2\sin x\cos x=0\cdot\cos x-(-1)\cdot\sin x;\\ 2\sin x\cos x=\sin x;\\ 2\sin x\cos x-\sin x=0;\\ \sin x(2\cos x-1)=0;\\ 1) \sin x=0\ \ \ x=\pi n, n\in Z;\\ 2) 2\cos x-1=0;\\ 2\cos x=1;\\ \cos x=\frac12;\\ x=\pm\arccos\frac12+2\pi k=\pm\frac\pi3+2\pi k. k\in Z\\ \left[ {{x=\pi n} \atop {x=\pm\frac\pi3+2\pi k}} \right.\ \ \ n,k\in Z$
2)
$x\in(\frac{4\pi}{3};4\pi]=(\pi+\frac{\pi}{3};4\pi]\\ 1) x=\pi n;\ \ \\ n=1:x=\pi\notin(\frac{4\pi}{3};4\pi];\\ n=2:x=2\pi\in(\frac{4\pi}{3};4\pi];\\ n=3:x=3\pi\in(\frac{4\pi}{3};4\pi];\\ n=4:x\in(\frac{4\pi}{3};4\pi];\\ n=5:x\notin(\frac{4\pi}{3};4\pi];\\$
$2) x=\pm\frac{\pi}{3}+2\pi k;\\ k=0:x=\pm\frac\pi3\notin(\frac{4\pi}{3};4\pi];\\ k=1: x=\pm\frac{\pi}{3}+2\pi= \left[ {{x=\frac{5\pi}{3}\in(\frac{4\pi}{3};4\pi];\\} \atop {x=\frac{7\pi}{3}\in(\frac{4\pi}{3};4\pi];\\}} \right. \\ k=2:x=\pm\frac\pi3+4\pi= \left[ {{x=\frac{11\pi}{3}\in(\frac{4\pi}{3};4\pi]} \atop {x=\frac{13\pi}{3}\notin(\frac{4\pi}{3};4\pi]}} \right.$
имеем такие ответы
$2) x=\pm\frac{\pi}{3}+2\pi k;\\ k=0:x=\pm\frac\pi3\notin(\frac{4\pi}{3};4\pi];\\ k=1: x=\pm\frac{\pi}{3}+2\pi= \left[ {{x=\frac{5\pi}{3}\in(\frac{4\pi}{3};4\pi];\\} \atop {x=\frac{7\pi}{3}\in(\frac{4\pi}{3};4\pi];\\}} \right. \\ k=2:x=\pm\frac\pi3+4\pi= \left[ {{x=\frac{11\pi}{3}\in(\frac{4\pi}{3};4\pi]} \atop {x=\frac{13\pi}{3}\notin(\frac{4\pi}{3};4\pi]}} \right. \\ x=\frac{5\pi}{3};\ \ 2\pi;\ \ \frac{7\pi}{3};\ \ 3\pi;\ \ \frac{11\pi}{3};\ \ 4\pi$