Question

-1+2cos^2+cos4x+√2 *cosx=0

Answer 1

$underbrace {-1+2cos^2x}_{cos2x}+cos4x+sqrt2cosx=0\\star ; ; cos2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x-1; star \\cos2x+cos4x+sqrt2cosx=0\\star ; ; cos2x+cos4x=2cos frac{2x+4x}{2}cdot cosfrac{4x-2x}{2}=2cos3xcdot cosx; star \\2cos3xcdot cosx+sqrt2cdot cosx=0\\cosx(2cos3x+sqrt2)=0\\1); cosx=0; ,; ; underline {x=frac{pi}{2}+pi n,; nin Z}\\2); ; cos3x=-frac{sqrt2}{2}; ,; ; 3x=pm arccos(-frac{sqrt2}{2})+2pi k,; kin Z \\3x=pm (pi -frac{pi}{4})+2pi k=pm frac{3pi}{4}+2pi k,; kin Z$

$underline {x=pm frac{pi}{4}+ frac{2pi k}{3}; ,; kin Z}$

Answer 2

огромнейшее спасибо :)