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Chapter 4: Problem 104

In Exercises \(99-104,\) find two values of \(\theta, 0 \leq \theta<2 \pi,\) that satisfy each equation. $$\tan \theta=-\frac{\sqrt{3}}{3}$$

Short Answer

Expert verified
The two values of \( \theta \) satisfying the equation \( \tan \theta = -\frac{\sqrt{3}}{3} \) are \( \theta = -\frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \)

Step by step solution

01

Apply Arctangent

To find the first value of \( \theta \), apply the arctangent function, or inverse tangent function, to both sides of the equation. This gives \( \theta = \arctan\left(-\frac{\sqrt{3}}{3}\right)\). Now determine the reference angle.
02

Determine the Reference Angle

The inverse of tangent of \(-\frac{\sqrt{3}}{3}\) gives -蟺/6. However, this is in the fourth quadrant, where tangent is negative. The reference angle equals 蟺/6, which corresponds to \(\frac{\sqrt{3}}{3}\) in a unit circle.
03

Find the Second Value of Theta

The tangent function has a period of 蟺, so to find the second value of \( \theta \), simply add 蟺 to the initial solution: \( \theta = -\frac{\pi}{6} + \pi = \frac{5\pi}{6} \). This is in the second quadrant, where tangent is also negative, fulfilling the condition of the original equation.

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