Question

решите тригонометрическое уравнение соs2х-соsх=0

Answer 1

$cos2x-cos x=0\ cos2x=2cos^2x-1\ 2cos^2x-1-cos x=0\ 2cos^2x-cos x-1=0\ cos x=t,;;cos^2x=t^2,;;tin[-1;1]\ 2t^2-t-1=0\ D=1+4cdot8=9\ t_1=1\ t_2=-frac12\ cos x=1Rightarrow x=2pi n,;ninmathbb{Z}\ cos x=-frac12Rightarrow x=frac{2pi}3+2pi n,;ninmathbb{Z}$

Answer 2

cos2x - cosx = 0
cos^2 x - sin^2 x - cosx = 0
cos^2 x - (1 - sin^2 x) - cosx = 0
2cos^2 x-1 - cosx = 0
cosx*(2cos x - 1)= 1
cos x = 1
x= 2*пи*n