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Answer:
[tex]\displaystyle x = \frac{\pi}{4} , \frac{5\pi}{4}[/tex]
Step-by-step explanation:
We want to find all solutions to the equation:
[tex]\displaystyle \sin 2x - 1 = 0[/tex]
In the interval [0, 2π).
First, add one to both sides:
[tex]\displaystyle \sin 2x = 1[/tex]
Recall that sin(u) equals one whenever u = π/2. This will occur for every rotation. Hence, we can say that u = π/2 + 2nπ where n is an integer.
And in this case, u = 2x. Thus:
[tex]\displaystyle 2x = \frac{\pi}{2} + 2n\pi\text{ where } n\in\mathbb{Z}[/tex]
Dividing both sides by two yields:
[tex]\displaystyle x = \frac{\pi}{4} + n\pi \text{ where } n\in \mathbb{Z}[/tex]
There are two values of x in the interval [0, 2π) given when n = 0 and n = 1:
[tex]\displaystyle x _ 1= \frac{\pi}{4} \text{ and } x_2 = \frac{\pi}{4} + (1)\pi = \frac{5\pi}{4}[/tex]
Any other solutions will be outside our interval.
Therefore, our solutions are:
[tex]\displaystyle x = \frac{\pi}{4} , \frac{5\pi}{4}[/tex]