Question

1)2cosx+5sinx=0
2)cos3x-cosx=0
3)sinx-5cosx=5
Какой нибудь примерчик
Это тригонометрия

Answer 1

$2cosx+5sinx=0|:cosx\\2+5tgx=0\\tgx=- \frac{2}{5} \\x=-arctg \frac{2}{5}+\pi n,n\in Z\\\\cos3x-cosx=0\\cos3x-cosx=-2 sin\frac{3x+x}{2}sin \frac{3x-x}{2}=-2sin2xsinx\\-2sin2xsinx=0\\sin2xsinx=0\\1)sin2x=0\\2x=\pi n\\x= \frac{\pi n}{2},n\in Z;\\2)sinx=0\\x=\pi n,n\in Z.$

$\\sinx-5cosx=5\\sinx=5+5cosx\\ \sqrt{1-cos^{2}x}=5+5cosx\\ (\sqrt{1-cos^{2}x})^{2}=(5+5cosx)^{2}\\ 1-cos^{2}x=25+50cosx+25cos^{2}x\\1-cos^{2}x-25-50cosx-25cos^{2}x=0\\-26cos^{2}x-50cosx-24=0|:(-2)\\ 13cos^{2}x+25cosx+12=0\\cosx=u\\13u^{2}+25u+12=0\\D:625-624=1\\u_1,_2= \frac{-25\pm1}{2*13}\\u_1= -\frac{12}{13}\\u_2=-1\\\\ 1)cosx= -\frac{12}{13}\\ x=\pm (arccos(- \frac{12}{13}))+2\pi n\\x=\pm (\pi-arccos \frac{12}{13})+2\pi n, n\in Z;\\\\2)cosx=-1\\x= \frac{\pi}{2} + \pi n, n\in Z$