How do I solve sqrt3 tan3theta + 1 = 0?

2 Answers
May 7, 2018

Within the interval 0 < theta < 2pi:

theta=(5pi)/18, (11pi)/18

Explanation:

.

sqrt3tan3theta+1=0

tan3theta=-1/sqrt3=-sqrt3/3

3theta=arctan(-sqrt3/3)

3theta=(5pi)/6,(11pi)/6

Within the interval 0 < theta < 2pi:

theta=(5pi)/18, (11pi)/18

May 7, 2018

theta= (5pi)/18+pi/3n

Explanation:

sqrt3 tan3theta + 1 = 0

Let u be 3theta

sqrt3 tanu + 1 = 0

sqrt3 tanu = -1

tanu = -1/sqrt3

tanu = -sqrt3/3

u= (5pi)/6, (11pi)/6

So:
3theta=(5pi)/6 +pin

theta= (5pi)/18+pi/3n
n is an element of all integers

graph{sqrt3tan(3x)+1 [-10, 10, -5, 5]}