fill in each blank so that the resulting statement is true. the solutions of sin x = √2/2 in 0,2π) are x =…

Explanation:

Step1: Recall sine - function properties

The sine function (y = \sin x) has a period of (2\pi). In the unit - circle, (\sin x) represents the (y) - coordinate of a point on the unit - circle corresponding to an angle (x). We know that (\sin x=\frac{\sqrt{2}}{2}) in the interval ([0,2\pi)). One solution is (x = \frac{\pi}{4}). The other solution in ([0,2\pi)) is (x=\frac{3\pi}{4}) since (\sin(\pi - \alpha)=\sin\alpha) and (\pi-\frac{\pi}{4}=\frac{3\pi}{4}).

Step2: Find the general solutions

Since the period of the sine function (y = \sin x) is (2\pi), the general solutions of the equation (\sin x=\frac{\sqrt{2}}{2}) are given by (x=\frac{\pi}{4}+2n\pi) and (x = \frac{3\pi}{4}+2n\pi), where (n\in\mathbb{Z}).

Answer:

The solutions of (\sin x=\frac{\sqrt{2}}{2}) in ([0,2\pi)) are (x = \frac{\pi}{4}) and (x=\frac{3\pi}{4}). If (n) is any integer, all solutions of (\sin x=\frac{\sqrt{2}}{2}) are given by (x=\frac{\pi}{4}+2n\pi) and (x=\frac{3\pi}{4}+2n\pi).