Question

решить:
1) cos (6x+1) = -корень из2/2
2) sinx=2cosx
3) ctg (2x- пи/10) = 1/корень из 3

Answer 1

$1); cos(6x+1)=-frac{sqrt2}{2}\6x+1=pm arccos(-frac{sqrt2}{2})+2pi n\6x+1=pmfrac{pi}{4}+2pi n\6x=pmfrac{pi}{4}-1+2pi n\x=pmfrac{pi}{24}-frac{1}{6}+frac{pi n}{6}, ; nin Z; \\2); sinx=2cosx|:cosx, ; cosx neq 0\tgx=2\x=arctg2+pi n, ; nin Z;\\3); ctg(2x-frac{pi}{10})=frac{1}{sqrt3}\2x-frac{pi}{10}=arcctgfrac{sqrt3}{3}+pi n\2x=frac{pi}{3}+frac{pi}{10}+pi n\x=frac{1}{2}*(frac{13pi}{30}+pi n)\x=frac{13pi}{60}+frac{pi n}{2}, ; nin Z.$