Question

помогите решить уравнение tg2x=1/корень 3
и sin (x-п/6)= 0

Answer 1

tg2x = 1/√3     2x = π/6 + πk       x = π/12 + πk/2     k∈Z

sin(x - π/6) = 0     x - π/6 = πn     x = π/6 + πn      n∈Z

Answer 2

$tg2x=frac{1}{sqrt{3} }\\2x=arctgfrac{1}{sqrt{3} }+pi n,nin z\\2x=frac{pi }{6}+pi n,nin z\\x=frac{pi }{12}+frac{pi n }{2},nin z$

$Sin(x-frac{pi }{6})=0\\x-frac{pi }{6}=pi n,nin z\\x=frac{pi }{6}+pi n,nin z$