Answer:
pi/6 and 11pi/6
Explanation
Given the trigonometry equation:
[tex]\begin{gathered} 3\text{sec}\theta\text{ - 2}\sqrt[]{3}=0 \\ \end{gathered}[/tex]
Add 2\sqrt[3] to both sides as shown;
[tex]\begin{gathered} 3\sec \theta-2\sqrt[]{3}=0+2\sqrt[]{3} \\ 3\sec \theta\text{ = 2}\sqrt[]{3} \\ \sec \text{ }\theta\text{ = }\frac{2\sqrt[]{3}}{3} \\ \frac{1}{\cos \theta}=\frac{2\sqrt[]{3}}{3} \\ \cos \theta\text{ =}\frac{3}{2\sqrt[]{3}} \\ \cos \text{ }\theta\text{ = }\frac{3\sqrt[]{3}}{2\cdot3} \\ \cos \text{ }\theta\text{ = }\frac{\sqrt[]{3}}{2} \\ \end{gathered}[/tex]
Take the cos inverse of both sides
[tex]\begin{gathered} \cos ^{-1}(\cos \theta)=cos^{-1}\frac{\sqrt[]{3}}{2} \\ \theta=cos^{-1}\frac{\sqrt[]{3}}{2} \\ \theta=30^0 \end{gathered}[/tex]
Since theta is between 0 and 2pi
theta = 360 - 30
theta = 330^0
Convert to radians
180^0 = pi rad
30^0 = x
180x = 30pi
x = 30pi/180
x = pi/6
Similarly;
180^0 = pi rad
330^0 = x
180x = 330pi
x = 330pi/180
x = 11pi/6
Hence the value of thets between 0 and 2pi are pi/6 and 11pi/6
