Question

Решите тригонометрию.
1.
$6cos{5x}*cos{7x}+frac{1}{3}=cos{2x}(8cos{4x}-1)+2cos{6x}$

Answer 1

$6cos{5x}*cos{7x}+frac{1}{3}=cos{2x}(8cos{4x}-1)+2cos{6x}\\6*frac{cos{12x}+cos{2x}}{2}+frac{1}{3}=8cos{2x}*cos{4x}-cos{2x}+2cos{6x}\\3cos{12x}+3cos{2x}+frac{1}{3}=8*frac{cos{6x}+cos{2x}}{2}-cos{2x}+2cos{6x}\\6cos^2{6x}-3+3cos{2x}+frac{1}{3}=4cos{6x}+3cos{2x}+2cos{6x};cos{6x}=a\6a^2-6a-frac{8}{3}=0|:2\3a^2-3a-frac{4}{3}=0;D=9+16=5^2\a=frac{3pm 5}{6}\begin{bmatrix}cos{6x}=-frac{1}{3}\cos{6x}=frac{8}{6}varnothingend{matrix}$

$Otvet:x=pm frac{1}{6}arccos{(frac{-1}{3})}+frac{pi n}{3}, nin Z.$