Question

решите уравнение cos3x+cosx+2sin^2x=1

Answer 1

$cos3x+cosx+2sin^2x=1\\2cdot cosfrac{3x+x}{2}cdot cosfrac{3x-x}{2}+2sin^2x=sin^2x+cos^2x\\2cdot cos2xcdot cosx+underbrace {sin^2x-cos^2x}_{-(cos^2x-sin^2x)}=0; ; ,; ; cos^2x-sin^2x=cos2x\\2cdot cos2xcdot cosx-cos2x=0\\cos2xcdot (2cosx-1)=0\\a); ; cos2x=0; ,; ; 2x=frac{pi}{2}+pi n; ; ,; ; x=frac{pi}{4}+frac{pi n}{2}; ,; nin Z\\b); ; cosx=frac{1}{2}; ,; ; x=pm frac{pi }{3}+2pi k; ,; kin Z\\Otvet:; ; x=frac{pi }{4}+frac{pi n}{2}; ,; x=pm frac{pi }{3}+2pi k; ,; ; n,kin Z$