# How do you solve the equation sqrt3sectheta+2=0 for -pi<=theta<=3pi?

Dec 21, 2016

$\frac{2 \pi}{3} , \frac{4 \pi}{3} , \frac{8 \pi}{3} , \frac{10 \pi}{3}$

#### Explanation:

$\sqrt{3.} \sec t + 2 = 0$.
$\frac{\sqrt{3}}{\cos t} = - 2$
$\sqrt{3} = - 2 \cos t$
$\cos t = - \frac{\sqrt{3}}{2}$

Trig table and unit circle -->
a. For interval $\left(- \pi , \pi\right)$ --> $\cos t = - \frac{\sqrt{3}}{2}$ --> arc $t = \pm \frac{2 \pi}{3}$
--> 2 solution arcs -->$t = \frac{2 \pi}{3}$ and $t = \frac{4 \pi}{3}$ (co-terminal)
b. For interval $\left(- \pi , \pi + 2 \pi\right)$ or interval $\left(- \pi , 3 \pi\right)$,
$t = \frac{2 \pi}{3} + 2 \pi = \frac{8 \pi}{3}$ and $t = \frac{4 \pi}{3} + 2 \pi = \frac{10 \pi}{3}$
Answers for $\left(- \pi , 3 \pi\right)$:
(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3)