Question

3tg6x+ корень из 3=0

Answer 1

$3\, tg\, 6x+\sqrt3=0\\\\tg\, 6x=-\frac{\sqrt3}{3}\\\\6x=-\frac{\pi}{6}+\pi n\; ,\; n\in Z\\\\x=-\frac{\pi }{36}+\frac{\pi n}{6}\; ,\; n\in Z$

Answer 2

I assume you're trying to solve the equation:

3tg(6x) + √3 = 0

To solve this equation, we need to isolate the variable, which is tg(6x). We can do this by subtracting √3 from both sides:

3tg(6x) = -√3

Next, we divide both sides by 3:

tg(6x) = -√3/3

Now, we can use the inverse tangent function to solve for 6x:

6x = arctan(-√3/3) + kπ (where k is an integer)

Finally, we divide both sides by 6 to get x:

x = [arctan(-√3/3) + kπ]/6

Therefore, the general solution to the equation 3tg(6x) + √3 = 0 is x = [arctan(-√3/3) + kπ]/6, where k is an integer.