Question

Помогите 3,4 задание

Answer 1

$3.\cos^2x=1-sin^2x=1-left(frac23right)^2=1-frac49=frac59\3cos^2x+2=3cdotfrac59+2=frac53+2=1frac23+2=3frac23\\\4.\;sinalpha=sqrt{1-cos^2alpha}=sqrt{1-0,64}=sqrt{0,36}=pm0,6\0 textgreater alpha textgreater fracpi2Rightarrowsin x textgreater 0\sin x=0,6\2ctgalpha=3cdotfrac{cosalpha}{sinalpha}=3cdotfrac{0,8}{0,6}=3cdotfrac43=4$