Question

Решите тригонометрию.
2.
$cos{5x}*cos{4x}+cos{4x}*cos{3x}=cos^2{2x}*cos{x}$

Answer 1

$cos{5x}*cos{4x}+cos{4x}*cos{3x}=cos^2{2x}*cos{x}\cos{4x}(cos{5x}+cos{3x})-cos^2{2x}*cos{x}=0\cos{4x}*2cos{4x}*cos{x}-cos^2{2x}*cos{x}=0\cos{x}(2cos^2{4x}-cos^2{2x})=0\begin{bmatrix}cos{x}=0\begin{matrix}2cos^2{4x}-cos^2{2x}=0&(1)end{matrix}end{matrix}\(1)2cos^2{4x}-frac{1+cos{4x}}{2}=0|*2\4cos^2{4x}-cos{4x}-1=0;D=1+16$

$begin{bmatrix}cos{4x}=frac{1-sqrt{17}}{8}begin{vmatrix}\\end{matrix}-sqrt{17}>-9\cos{4x}=frac{1+sqrt{17}}{8}begin{vmatrix}\\end{matrix}sqrt{17}<7end{matrix}\Otvet:x=begin{Bmatrix}pm frac{1}{4}arccos{(frac{1pm sqrt{17}}{8})}+frac{pi n}{2};pm frac{pi }{2}+2pi nend{Bmatrix},nin Z.$