To solve the equation [tex]\(\sin(2x) - \cos(x) = 0\)[/tex], follow these steps:
1. Express [tex]\(\sin(2x)\)[/tex] in terms of [tex]\(\sin(x)\)[/tex] and [tex]\(\cos(x)\)[/tex]:
The double-angle identity for sine is [tex]\(\sin(2x) = 2\sin(x)\cos(x)\)[/tex]. Using this identity, the original equation becomes:
[tex]\[
2\sin(x)\cos(x) - \cos(x) = 0
\][/tex]
2. Factor out the common term [tex]\(\cos(x)\)[/tex]:
[tex]\[
\cos(x) (2\sin(x) - 1) = 0
\][/tex]
This equation is satisfied if either [tex]\(\cos(x) = 0\)[/tex] or [tex]\(2\sin(x) - 1 = 0\)[/tex].
3. Solve for [tex]\(x\)[/tex] when [tex]\(\cos(x) = 0\)[/tex]:
The cosine function is zero at odd multiples of [tex]\(\frac{\pi}{2}\)[/tex]:
[tex]\[
x = \frac{\pi}{2} + k\pi \quad \text{for any integer } k
\][/tex]
By considering one period [tex]\([0, 2\pi)\)[/tex] as a principal range, the solutions relevant to this range are:
[tex]\[
x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}
\][/tex]
4. Solve for [tex]\(x\)[/tex] when [tex]\(2\sin(x) - 1 = 0\)[/tex]:
Isolate [tex]\(\sin(x)\)[/tex]:
[tex]\[
2\sin(x) = 1 \implies \sin(x) = \frac{1}{2}
\][/tex]
The sine function equals [tex]\(\frac{1}{2}\)[/tex] at:
[tex]\[
x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \quad \text{for any integer } k
\][/tex]
Again, restricting to one period [tex]\([0, 2\pi)\)[/tex], the solutions are:
[tex]\[
x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}
\][/tex]
Combining all the solutions we found within one period [tex]\([0, 2\pi)\)[/tex], we have:
[tex]\[
x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}
\][/tex]
Since we are generally interested in the principal values, including standard intervals, a simplified list would be:
[tex]\[
x = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}
\][/tex]
Thus, the complete set of solutions to the equation [tex]\(\sin(2x) - \cos(x) = 0\)[/tex] includes the following angles:
[tex]\[
x = -\frac{\pi}{2}, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}
\][/tex]
