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