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}
]
