Question

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

Answer 1

$\pm\frac{\pi }{9}+\frac{2\pi m }{3} ,~m \in\mathbb {Z} ; \frac{\pi }{2} +\pi n, ~n\in\mathbb {Z}.$

Объяснение:

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

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

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