Question

ПОМОГИТЕ ПОЖАЛУЙСТААААААА!!!!!!!!!
1-cos2x+3cosx=0

Answer 1

$1-cos2x+3cosx=0\\sin^2x=frac{1-cos2x}{2}; ; to ; ; 1-cos2x=2sin^2x=2(1-cos^2x)=2-2cos^2x\\2-2cos^2x+3cosx=0\\t=cosx; ,; ; -1 leq cosx leq 1\\2t^2-3t-2=0\\D=9+16=25; ,; t_1=frac{3-5}{4}=-frac{1}{2}; ,; ; t_2=2 textgreater 1\\cosx=-frac{1}{2}; ,\\ x=pm (pi -arccos frac{1}{2})+2pi n=pm (pi -frac{pi}{3})+2pi n=pm frac{2pi}{3}+2pi n,; nin Z$