Question

Найдите корни уравнения::::::

Answer 1

$1)15Cos(\frac{\pi }{2}-5x)=0\\\\15Sin5x=0\\\\Sin5x=0\\\\5x=\pi n,n\in Z\\\\x=\frac{\pi n }{5},n\in Z\\\\n=-1\\x=-\frac{\pi }{5}\\\\n=0\\\\x=0\\\\n=1\\\\x=\frac{\pi }{5}\\\\n=2\\\\x=\frac{2\pi }{5} \\\\n=3\\\\x=\frac{3\pi }{5}\\\\n=4\\\\x=\frac{4\pi }{5} \\\\n=5\\\\x=\pi$

$2)Sin2xCos2x=-\frac{1}{4}\\\\2Sin2xCos2x=-\frac{1}{4}*2\\\\Sin4x=-\frac{1}{2}\\\\1)4x=-\frac{\pi }{6}+2\pi n,\in Z \\\\x=-\frac{\pi }{24}+\frac{\pi n }{2},n\in Z\\\\2)4x=-\frac{5\pi }{6}+2\pi n,n\in Z\\\\x=-\frac{5\pi }{24}+\frac{\pi n }{2},n\in Z\\\\n=0\\\\x=-\frac{\pi }{24}\\\\x=-\frac{5\pi }{24}\\\\n=1\\\\x=-\frac{5\pi }{24}+\frac{\pi }{2}=\frac{7\pi }{24}$

$3)Cos(\frac{\pi }{2}-\frac{2x}{3})=Sin\frac{\pi }{3}\\\\Cos\frac{2x}{3}=\frac{\sqrt{3} }{2}\\\\1)\frac{2x}{3}=\frac{\pi }{6} +2\pi n,n\in Z\\\\x=\frac{\pi }{4}+3\pi n,n\in Z\\\\2)\frac{2x}{3}=-\frac{\pi }{6}+2\pi n,n\in Z\\\\x=-\frac{\pi }{4}+3\pi n,n\in Z$

В этом промежутке корней нет