Question

3 tg (2x + pi/4) + √3 = 0

Answer 1

$mathtt{3tg(2x+frac{pi}{4})+sqrt{3}=0;~tg(2x+frac{pi}{4})=-frac{sqrt{3}}{3}}$

$mathtt{2x+frac{pi}{4}=-frac{pi}{6}+pi n,nin Z}$

следовательно, наш корень $mathtt{x=frac{12npi-5pi}{24},nin Z}$