Solve the equation in the interval $$ 0 \leq \theta

Question

Solve the equation in the interval $$ 0 \leq \theta<2 \pi $$ $$ \sin (4 \theta)-\sin (2 \theta)=0 $$ What are the solutions in the interval $$ 0 \leq \theta<2 \pi $$ ? A. $$ \left\{0, \frac{\pi}{3}, \frac{\pi}{2}, \frac{5 \pi}{6}, \pi, \frac{4 \pi}{3}, \frac{3 \pi}{2}, \frac{11 \pi}{6}\right\} $$ B. $$ \left\{0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \pi, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}\right\} $$ C. $$ \left\{0, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{3 \pi}{2}, \frac{5 \pi}{3}\right\} $$ D. $$ \left\{0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{2 \pi}{3}, \pi, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{5 \pi}{3}\right\} $$ E. $$ \left\{0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\} $$ F. $$ \left\{0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\} $$

UpStudy AI · Accepted Answer

To solve the equation

$$
\sin(4\theta) - \sin(2\theta) = 0
$$

we can use the identity that states $$ \sin A - \sin B = 0 $$ implies $$ A = B + 2k\pi $$ or $$ A = \pi - B + 2k\pi $$ for some integer $$ k $$.

### Step 1: Set up the equations

From the equation, we have:

1. $$ 4\theta = 2\theta + 2k\pi $$
2. $$ 4\theta = \pi - 2\theta + 2k\pi $$

### Step 2: Solve the first equation

From the first equation:

$$
4\theta = 2\theta + 2k\pi
$$

Subtract $$ 2\theta $$ from both sides:

$$
2\theta = 2k\pi
$$

Dividing by 2 gives:

$$
\theta = k\pi
$$

For $$ k = 0 $$: $$ \theta = 0 $$

For $$ k = 1 $$: $$ \theta = \pi $$

For $$ k = 2 $$: $$ \theta = 2\pi $$ (not in the interval $$ [0, 2\pi) $$)

Thus, from the first equation, we have the solutions $$ \theta = 0, \pi $$.

### Step 3: Solve the second equation

From the second equation:

$$
4\theta = \pi - 2\theta + 2k\pi
$$

Rearranging gives:

$$
4\theta + 2\theta = \pi + 2k\pi
$$

This simplifies to:

$$
6\theta = \pi(1 + 2k)
$$

Dividing by 6 gives:

$$
\theta = \frac{\pi(1 + 2k)}{6}
$$

### Step 4: Find values of $$ k $$

Now we will find values of $$ k $$ such that $$ 0 \leq \theta < 2\pi $$:

1. For $$ k = 0 $$: 
   $$
   \theta = \frac{\pi}{6}
   $$
2. For $$ k = 1 $$: 
   $$
   \theta = \frac{\pi(1 + 2 \cdot 1)}{6} = \frac{3\pi}{6} = \frac{\pi}{2}
   $$
3. For $$ k = 2 $$: 
   $$
   \theta = \frac{\pi(1 + 2 \cdot 2)}{6} = \frac{5\pi}{6}
   $$
4. For $$ k = 3 $$: 
   $$
   \theta = \frac{\pi(1 + 2 \cdot 3)}{6} = \frac{7\pi}{6}
   $$
5. For $$ k = 4 $$: 
   $$
   \theta = \frac{\pi(1 + 2 \cdot 4)}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}
   $$
6. For $$ k = 5 $$: 
   $$
   \theta = \frac{\pi(1 + 2 \cdot 5)}{6} = \frac{11\pi}{6}
   $$

### Step 5: Compile all solutions

The solutions we have found are:

- From the first equation: $$ \theta = 0, \pi $$
- From the second equation: $$ \theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} $$

Combining all these, the complete set of solutions in the interval $$ 0 \leq \theta < 2\pi $$ is:

$$
\left\{0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}
$$

### Conclusion

The correct answer is:

**B.** $$ \left\{0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\} $$
The solutions are $$ \left\{0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\} $$, which corresponds to choice B.

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