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

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

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