
Homework:8.6 HW - Double-angle and Half-angle Formulas
Question 25, 8.6.81
HW Score: 70.31%, 22.5 of 32 points
 Points: 0 of 1


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Part 1
Solve the equation on the interval 0less than or equalsthetaless than2pi.
tangent left parenthesis 2 theta right parenthesis plus 2 sine theta equals 0
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes in your choice.
A.
The solution set is StartSet nothing EndSet
  
.
​(Use a comma to separate answers as needed. Type an exact​ answer, using pi as needed. Use integers or fractions for any numbers in the​ expression.) 
B.
There is no solution.
tangent left parenthesis 2 theta right parenthesis plus 2 sine theta equals 0Select the correct choice below and fill in any answer boxes in your choice.
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We are tasked with solving the trigonometric equation:

\[
\text{tan}(2\theta) + 2\sin(\theta) = 0
\]

on the interval \(0 \leq \theta < 2\pi\).

### Step 1: Rewrite the equation in terms of known identities
Recall the double-angle identity for tangent:
\[
\text{tan}(2\theta) = \frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)}
\]
So the equation becomes:
\[
\frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)} + 2\sin(\theta) = 0
\]

### Step 2: Simplify
Factor out \( \sin(\theta) \) from both terms:
\[
\sin(\theta) \left(\frac{2\cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)} + 2\right) = 0
\]

This gives two cases to consider:

### Case 1: \( \sin(\theta) = 0 \)
For this, \( \theta = 0, \pi, 2\pi \), since sine is zero at these points. But we need to check these values for validity later.

### Case 2: Solve the remaining equation
For the remaining part:
\[
\frac{2\cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)} + 2 = 0
\]
Multiply through by \( \cos^2(\theta) - \sin^2(\theta) \) to simplify:
\[
2\cos(\theta) + 2(\cos^2(\theta) - \sin^2(\theta)) = 0
\]
Simplifying further:
\[
2\cos(\theta) + 2\cos(2\theta) = 0
\]
Divide through by 2:
\[
\cos(\theta) + \cos(2\theta) = 0
\]

### Step 3: Solve \( \cos(\theta) + \cos(2\theta) = 0 \)
Using the double angle identity for cosine, \( \cos(2\theta) = 2\cos^2(\theta) - 1 \), we get:
\[
\cos(\theta) + 2\cos^2(\theta) - 1 = 0
\]
Let \( x = \cos(\theta) \), and rewrite the equation:
\[
x + 2x^2 - 1 = 0
\]
This is a quadratic equation:
\[
2x^2 + x - 1 = 0
\]
Solve this using the quadratic formula:
\[
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}
\]
So,
\[
x = \frac{-1 + 3}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-1 - 3}{4} = -1
\]

### Step 4: Solve for \( \theta \)
Now we solve for \( \theta \) using these values of \( x = \cos(\theta) \).

- If \( \cos(\theta) = \frac{1}{2} \), then \( \theta = \frac{\pi}{3}, \frac{5\pi}{3} \).
- If \( \cos(\theta) = -1 \), then \( \theta = \pi \).

### Step 5: Combine solutions
We already found that \( \sin(\theta) = 0 \) gives \( \theta = 0, \pi, 2\pi \). We now add the solutions from \( \cos(\theta) = \frac{1}{2} \) and \( \cos(\theta) = -1 \), which are \( \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \).

Thus, the solution set is:
\[
\theta = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi
\]

Would you like more details on any step? Here are some related questions to deepen your understanding:

1. How do double-angle identities help solve trigonometric equations?
2. What are the key steps in solving quadratic equations in trigonometry?
3. Why do we check both sine and cosine values when solving trigonometric equations?
4. Can we apply a different identity for \( \tan(2\theta) \)?
5. How would the solutions change if the interval for \( \theta \) were different?

**Tip:** Always verify your solutions by plugging them back into the original equation to check for extraneous solutions.

Solve the equation on the interval 0 ≤ θ < 2π: tangent(2θ) + 2sin(θ) = 0. Select the correct choice below and fill in any answer boxes in your choice. A) The solution set is { } or B) There is no solution.

Explore how to solve the trigonometric equation tan(2θ) + 2sin(θ) = 0 over the interval 0 ≤ θ < 2π. This step-by-step guide applies double-angle identities and quadratic equations, ideal for students in Grades 11-12. Learn how to break down complex trigonometric functions to find exact solutions.

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