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Chapter 4: Problem 103

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

Short Answer

Expert verified
The two values of \(\theta, 0 \leq \theta<2 \pi,\) that satisfy the equation \(\tan \theta=-\sqrt{3}\) are \(2\pi /3\) and \(4\pi /3\).

Step by step solution

01

Identify the related acute angle

Let's firstly find the related acute angle where \(\tan\theta = \sqrt{3}\). By using special right triangles or the unit circle, we know that this happens when \(\theta=\pi /3\) or \(60^\circ \). This angle is in the first quadrant, but we are looking for angles where \(\tan\theta\) is negative.
02

Find the angle in the second quadrant

In the second quadrant, the \(\tan\) function is negative. The angle that has the same \(\tan\) value as our related angle but is in the second quadrant can be found by subtracting our related angle from \(\pi\). \(\pi - \pi/3 = 2\pi /3\). So one solution is \(2\pi /3\).
03

Find the angle in the fourth quadrant

Similarly, in the fourth quadrant, the \(\tan\) function is negative. The angle in the fourth quadrant that has the same \(\tan\) value as our related angle can be found by adding our related angle to \(\pi\). \(\pi + \pi/3 = 4\pi /3\). So the other solution is \(4\pi /3\).
04

Confirm the solutions

We can confirm our solutions by plugging them back into the original equation. If the left side of the equation equals the right side (i.e., \(-\sqrt{3}\)), then our solutions are correct.

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