Question

Решите пожалуйста срочно очень прошу

Answer 1

$1.\frac{7pi}6=frac{7cdot180}{6}=7cdot30=210^o\frac{11pi}5=frac{11cdot180}5=11cdot36=396^o\\2.\330^o=frac{330pi}{180}=frac{11pi}6\\120^o=frac{120pi}{180}=frac{2pi}3\\3.;160^o;-;II;4ETB.,sin160^o>0\tgleft(-frac{13pi}6right)=-tg(390^o)<0\\4.;cos765^o=cosfrac{17pi}4=cos(2pi+fracpi4)=cosfracpi4=frac1{sqrt2}\ctgfrac{19pi}4=ctgleft(4pi+frac{3pi}4right)=ctgfrac{3pi}4=-1$

$5.;3tgfracpi5cdot ctgfracpi5+2sinfracpi3=3cdot1+3cdotfrac{sqrt3}2=3+frac{3sqrt3}2\\6.;frac{2sinleft(fracpi2-alpharight)}{5cos(pi+alpha)}=frac{2cosalpha}{-5cosalpha}-frac25\\7.;2cos x+sqrt3=0\cos x=-frac{sqrt3}2\x=pmfrac{5pi}6+pi n,;ninmathbb{Z}\\2sinleft(fracpi3-xright)=1\sinleft(fracpi3-xright)=frac12\fracpi3-x=(-1)^ncdotfracpi6+pi n\x=fracpi3-(-1)^ncdotfracpi6-pi n,;ninmathbb{Z}$