Question

СРОЧНО С1!!88 баллов даю!!!!

Answer 1

Думаю, можно разобраться...

Answer 2

a)$16cos^4x-24cos^2x+9=0;\ cos x=t, -1leq xleq1;\ 16t^4-24t^2+9=0;\ (4t^2)^2-2cdot4t^2cdot3+3^2=0;\ (4t^2-3)=0;\ 4t^2-3=0;\ 4t^2=3;\ t^2=frac{3}{4};\ t=pmfrac{sqrt3}{2};$
$cos x=pmfrac{sqrt3}{2};\ cos x=frac{sqrt3}{2};\ x=pmarccosfrac{sqrt3}{2}+2pi n, nin Z;\ x=pmfrac{pi}{6}+2pi n, nin Z;\ cos x=-frac{sqrt3}{2};\ x=pmarccos(-frac{sqrt3}{2})+2pi k, k in Z;\ x=pmfrac{5pi}{6}+2pi k;\ left[ {{x=pmfrac{pi}{6}+2pi n,} atop {x=pipmfrac{pi}{6}+2pi k}} right. n,kin Z$


б)
$xin [2pi,3pi];\ 2pi+frac{pi}{6}=frac{12pi+pi}{6}=frac{13pi}{6};\ 2pi+pi-frac{pi}{6}=3pi-frac{pi}{6}=frac{18pi-pi}{6}=frac{17pi}{6};\ \ \ \ x=frac{13pi}{6};\ x=frac{17pi}{6};$