Question

Find all solutions of the equation 2 sine x cosine 2 x minus cosine 2 x equals 0 over the interval 0 less-than x less-than-or-equal-to pi.
(1 point)
Responses

x equals start fraction pi over 6 end fraction comma start fraction pi over 2 end fraction comma start fraction 5 pi over 6 end fraction comma pi
Image with alt text: x equals start fraction pi over 6 end fraction comma start fraction pi over 2 end fraction comma start fraction 5 pi over 6 end fraction comma pi

x equals start fraction pi over 6 end fraction comma start fraction pi over 4 end fraction comma start fraction 3 pi over 4 end fraction comma start fraction 5 pi over 6 end fraction
Image with alt text: x equals start fraction pi over 6 end fraction comma start fraction pi over 4 end fraction comma start fraction 3 pi over 4 end fraction comma start fraction 5 pi over 6 end fraction

x equals start fraction pi over 4 end fraction comma start fraction pi over 3 end fraction comma start fraction 2 pi over 3 end fraction comma start fraction 3 pi over 4 end fraction
Image with alt text: x equals start fraction pi over 4 end fraction comma start fraction pi over 3 end fraction comma start fraction 2 pi over 3 end fraction comma start fraction 3 pi over 4 end fraction

x equals start fraction pi over 3 end fraction comma start fraction pi over 2 end fraction comma start fraction 2 pi over 3 end fraction comma pi
Image with alt text: x equals start fraction pi over 3 end fraction comma start fraction pi over 2 end fraction comma start fraction 2 pi over 3 end fraction comma pi

Answer 1

To solve the equation

\[ 2 \sin x \cos 2x - \cos 2x = 0, \]

we can factor out \(\cos 2x\):

\[ \cos 2x (2 \sin x - 1) = 0. \]

This gives us two cases to solve:

1. \( \cos 2x = 0 \)
2. \( 2 \sin x - 1 = 0 \)

### Solving \( \cos 2x = 0 \)

The cosine function equals zero at odd multiples of \(\frac{\pi}{2}\):

\[
2x = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z}.
\]

Dividing by 2, we get:

\[
x = \frac{\pi}{4} + \frac{k\pi}{2}.
\]

We need to find solutions in the interval \( 0 < x \leq \pi \):

- For \( k = 0 \):
\[
x = \frac{\pi}{4} \quad (\text{valid})
\]
- For \( k = 1 \):
\[
x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \quad (\text{valid})
\]
- For \( k = 2 \):
\[
x = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \quad (\text{not valid})
\]

Thus, from the equation \( \cos 2x = 0 \), we have the solutions:

\[
x = \frac{\pi}{4}, \frac{3\pi}{4}.
\]

### Solving \( 2 \sin x - 1 = 0 \)

From this equation, we get:

\[
\sin x = \frac{1}{2}.
\]

The sine function equals \(\frac{1}{2}\) at:

\[
x = \frac{\pi}{6}, \frac{5\pi}{6}.
\]

### Summary of Solutions

Combining all the values we found, we have:

1. From \( \cos 2x = 0\):
- \( x = \frac{\pi}{4}, \frac{3\pi}{4} \)
2. From \( 2 \sin x - 1 = 0 \):
- \( x = \frac{\pi}{6}, \frac{5\pi}{6} \)

Thus, the complete set of solutions in the interval \(0 < x \leq \pi\) is:

\[
x = \frac{\pi}{6}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{6}.
\]

Among the provided response options, the correct one is:

**x equals start fraction pi over 6 end fraction comma start fraction pi over 4 end fraction comma start fraction 3 pi over 4 end fraction comma start fraction 5 pi over 6 end fraction.**