Sin2x -coax=0

To solve the equation \( \sin(2x) - \cos(x) = 0 \), we can rearrange it to isolate one of the trigonometric functions:

\[ \sin(2x) = \cos(x) \]

Next, we can use the double angle identity for sine:

\[ \sin(2x) = 2\sin(x)\cos(x) \]

This gives us:

\[ 2\sin(x)\cos(x) = \cos(x) \]

Assuming \( \cos(x) \neq 0 \), we can divide both sides by \( \cos(x) \):

\[ 2\sin(x) = 1 \]

This simplifies to:

\[ \sin(x) = \frac{1}{2} \]

The solutions for \( x \) corresponding to \( \sin(x) = \frac{1}{2} \) are:

\[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]

Now we must also consider the case where \( \cos(x) = 0 \). This occurs when:

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

Putting everything together, the complete solution set for the equation \( \sin(2x) - \cos(x) = 0 \) is:

\[ x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad \text{and} \quad x = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z} \]