Question

решите уравнение cos2x+cos4x+2sin^2x/2=1

Answer 1

Ответ:

$pmfrac{pi }{9}+frac{2pi m }{3} ,~m inmathbb {Z} ; frac{pi }{2} +pi  n, ~ninmathbb {Z}.$

Объяснение:

$cos2x+cos4x+2sin^{2} frac{x}{2} =1;\cos2x+cos4x=1-2sin^{2} frac{x}{2};\\2 cos frac{2x+4x}{2} *cosfrac{2x-4x}{2} =cos(2*frac{x}{2} );\\2cos3x*cos(-x) =cosx;\2cos3x*cosx-cosx=0;\cosx( 2cos3x-1)=0;$

$left [begin{array}{lcl} {{cosx=0,} \ {2cos3x-1=0;}} end{array} right.Leftrightarrow left [ begin{array}{lcl} {{x=frac{pi }{2} +pi n ,~ninmathbb {Z}}  \ {cos3x=frac{1}{2}; }} end{array} right.Leftrightarrow$

$left [ begin{array}{lcl} {{x=frac{pi }{2} +pi n, ~ninmathbb {Z}  } \ {3x=pmfrac{pi }{3}+2pi  m,~m inmathbb {Z} }} end{array} right. Leftrightarrow left [ begin{array}{lcl} {{x=frac{pi }{2} +pi n, ~ninmathbb {Z}  } \ {x=pmfrac{pi }{9}+frac{2pi m }{3} ,~m inmathbb {Z} }} end{array} right.$