Question

Solve the trig equation on the interval
`0 \leqslant theta \: \ \textless \ 2\pi`
`3 sec \: \: theta \: - 2 √(3 ) = 0`

Answer 1

pi/6 and 11pi/6

Step-by-step explanation

Given the trigonometry equation:

`\begin{gathered} 3\text{sec}\theta\text{ - 2}\sqrt[]{3}=0 \\ \end{gathered}`

Add 2\sqrt[3] to both sides as shown;

`\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} \\ (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}`

Take the cos inverse of both sides

`\begin{gathered} \cos ^(-1)(\cos \theta)=cos^(-1)\frac{\sqrt[]{3}}{2} \\ \theta=cos^(-1)\frac{\sqrt[]{3}}{2} \\ \theta=30^0 \end{gathered}`

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