Math  /  Trigonometry

QuestionDetermine the specific solutions (if any) on the interval [0,2π[0,2 \pi ). tanθ+3=0\tan \theta+\sqrt{3}=0
Select the correct choice below and, if necessary, fill in the answer box withit A. θ=\theta= \square (Use a comma to separate answers as needed. Simplify your answer B. There is no solution.

Studdy Solution

STEP 1

What is this asking? Find all angles θ\theta between 00 and 2π2\pi (inclusive of 00 and exclusive of 2π2\pi) where tanθ+3=0\tan \theta + \sqrt{3} = 0. Watch out! Remember tanθ\tan \theta has a period of π\pi, so there might be multiple solutions within the given interval!
Also, be careful with the signs.

STEP 2

1. Isolate the tangent function
2. Find the principal solution
3. Find all solutions in the given interval

STEP 3

Let's **isolate** tanθ\tan \theta!
We start with our equation: tanθ+3=0 \tan \theta + \sqrt{3} = 0 We want to get tanθ\tan \theta by itself, so we'll **subtract** 3\sqrt{3} from both sides of the equation: tanθ+33=03 \tan \theta + \sqrt{3} - \sqrt{3} = 0 - \sqrt{3} tanθ=3 \tan \theta = -\sqrt{3} Great! Now we have tanθ\tan \theta isolated.

STEP 4

We're looking for an angle whose tangent is 3-\sqrt{3}.
We know that tanπ3=3\tan \frac{\pi}{3} = \sqrt{3}.
Since the tangent function is negative in the second and fourth quadrants, we need to find angles in those quadrants related to π3\frac{\pi}{3}.

STEP 5

In the second quadrant, the angle is ππ3=2π3\pi - \frac{\pi}{3} = \frac{2\pi}{3}.
So, tan2π3=3\tan \frac{2\pi}{3} = -\sqrt{3}.
This is our **principal solution**!

STEP 6

Since the tangent function has a period of π\pi, we can find other solutions by **adding** multiples of π\pi to our principal solution.

STEP 7

Our principal solution is 2π3\frac{2\pi}{3}.
Let's add π\pi to it: 2π3+π=2π3+3π3=5π3 \frac{2\pi}{3} + \pi = \frac{2\pi}{3} + \frac{3\pi}{3} = \frac{5\pi}{3} This is still within our interval of [0,2π)[0, 2\pi).

STEP 8

If we add another π\pi to 5π3\frac{5\pi}{3}, we get 5π3+π=8π3\frac{5\pi}{3} + \pi = \frac{8\pi}{3}, which is greater than 2π2\pi, so it's outside our interval.

STEP 9

Therefore, our solutions within the interval [0,2π)[0, 2\pi) are 2π3\frac{2\pi}{3} and 5π3\frac{5\pi}{3}.

STEP 10

A. θ=2π3,5π3\theta = \frac{2\pi}{3}, \frac{5\pi}{3}

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