To find all solutions of the equation [tex]\( 2 \sin \theta - \sqrt{2} = 0 \)[/tex] within the interval [tex]\( [0, 2\pi) \)[/tex], follow these steps:
1. Isolate the sine function:
[tex]\(2 \sin \theta - \sqrt{2} = 0\)[/tex]
Add [tex]\(\sqrt{2}\)[/tex] to both sides:
[tex]\(2 \sin \theta = \sqrt{2}\)[/tex]
Divide both sides by 2:
[tex]\(\sin \theta = \frac{\sqrt{2}}{2}\)[/tex]
2. Recognize standard values:
The value [tex]\(\frac{\sqrt{2}}{2}\)[/tex] is a known value of the sine function. From the unit circle, we know that:
[tex]\(\sin \theta = \frac{\sqrt{2}}{2}\)[/tex] at:
[tex]\(\theta = \frac{\pi}{4}\)[/tex] (for the first quadrant) and
[tex]\(\theta = \frac{3\pi}{4}\)[/tex] (for the second quadrant).
3. State the solutions:
Therefore, the solutions of the equation [tex]\( 2 \sin \theta - \sqrt{2} = 0 \)[/tex] within the interval [tex]\( [0, 2\pi) \)[/tex] are:
[tex]\(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)[/tex]
So, the solutions in radians are:
[tex]\[ \boxed{\frac{\pi}{4}, \frac{3\pi}{4}} \][/tex]
