Solve: [tex]2sin(2x) = 1[/tex]
Divide both sides by 2.
[tex]sin(2x) = \frac{1}{2}[/tex]
Take the sine inverse of both sides:
[tex]2x = sin^{-1}(\frac{1}{2})[/tex]
[tex]2x = \frac{\pi}{6}, \frac{5\pi}{6}, 0 \leq x \leq 2\pi[/tex]
[tex]x = \frac{\pi}{12}, \frac{5\pi}{12}[/tex]
But we know there are more solutions if we extend the domain. In fact, there are infinitely more solutions since the domain of sine is all real x values.
Thus, we can develop a general solution:
For every [tex]\pi[/tex] units, there is another solution for both [tex]\frac{\pi}{12}[/tex] and [tex]\frac{5\pi}{12}[/tex]
General solutions:
[tex]x = \pi n + \frac{\pi}{12}, n \in \mathbb{Z}[/tex]
[tex]x = \pi n + \frac{5\pi}{12}, n \in \mathbb{Z}[/tex]
