# How do you solve tan(2x + pi/3) = 1 from 0 to 2pi?

Nov 6, 2015

(11pi)/24 ; (23pi)/24 ; (33pi)/24 , (35pi)/24

#### Explanation:

Trig Table and unit circle -->
$\tan \left(\frac{\pi}{4}\right) = 1$ and $\tan \left(\frac{\pi}{4} + \pi\right) = \tan \left(\frac{5 \pi}{4}\right) = 1$
Substitute in the right side of the equation 1 by tan (pi/4), and 1 by tan ((5pi)/4):

a. $\tan \left(2 x + \frac{\pi}{3}\right) = 1 = \tan \left(\frac{\pi}{4}\right)$
$2 x + \frac{\pi}{3} = \frac{\pi}{4} + k \pi$
$2 x = \frac{\pi}{4} - \frac{\pi}{3} = - \frac{\pi}{12} + k \pi$ --> $x 1 = - \frac{\pi}{24} + \frac{k \pi}{2}$
$x 1 = - \frac{\pi}{24} = \left(\frac{23 \pi}{24}\right) + \frac{k \pi}{2}$ (co-terminal arc)
If k = 1 --> $x 2 = \frac{23 \pi}{24} + \frac{\pi}{2} = \frac{35 \pi}{24}$
b. $\tan \left(2 x + \frac{\pi}{3}\right) = \tan \left(\frac{5 \pi}{4}\right)$
$2 x + \frac{\pi}{3} = \frac{5 \pi}{4} + k \pi$
$2 x = \frac{5 \pi}{4} - \frac{\pi}{3} = \frac{11 \pi}{12} + k \pi$ --> $x 3 = \frac{11 \pi}{24} + \frac{k \pi}{2}$
If k = 1 --> $x 4 = \frac{11 \pi}{24} + \frac{\pi}{2} = \frac{33 \pi}{24}$
Answers for $\left(0 , 2 \pi\right)$:
$\frac{11 \pi}{24} , \frac{23 \pi}{24} , \frac{33 \pi}{24} , \frac{35 \pi}{24}$