Question

решите пожалуйста очень срочно нужно

Answer 1

$3tg(4x-\frac{\pi }{6})-\sqrt{3} =0\\\\3tg(4x-\frac{\pi }{6})=\sqrt{3}\\\\tg(4x-\frac{\pi }{6})=\frac{\sqrt{3} }{3}\\\\4x-\frac{\pi }{6}=arctg\frac{\sqrt{3} }{3}+\pi n,n\in Z\\\\4x-\frac{\pi }{6}=\frac{\pi }{6}+\pi n,n\in Z\\\\4x=\frac{\pi }{6}+\frac{\pi }{6}+\pi n,n\in Z\\\\4x=\frac{\pi }{3}+\pi n,n\in Z\\\\x=\frac{\pi }{12}+\frac{\pi n }{4},n\in Z$

Answer 2

3tg(4x - (π/6)) - √3 = 0

Пусть 4x - (π/6) = a, тогда 3tg(a) = √3 ⇒ tg(a) = √3/3 ⇒

a = (π/3) + πn

4x - (π/6) = (π/6) + πn

4x = (π/6) + (π/6) + πn

4x = (π/3) + πn

x = (π/12) + (πn/4), n ∈ Z

ОТВЕТ: (π/12) + (πn/4) , n ∈ Z