Given:
[tex]\sin\theta=-\frac{\sqrt{2}}{2},0\leq\theta<2\pi[/tex]To find:
The two values.
Explanation:
It can be written as,
[tex]\begin{gathered} \sin\theta=-\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{\sqrt{2}} \\ \sin\theta=-\frac{2}{2\sqrt{2}} \\ \sin\theta=-\frac{1}{\sqrt{2}} \\ \theta=\sin^{-1}(-\frac{1}{\sqrt{2}}) \end{gathered}[/tex]Since the angle lies between 0 and 2π.
Also, the value of the sine function is negative only when the angle lies in the third and fourth quadrant.
So, the angles are,
[tex]\begin{gathered} \theta=180+45^{\circ}\Rightarrow\text{ }225^{\circ}\text{ }(or)\text{ }\frac{5\pi}{4} \\ \theta=360-45\Rightarrow\text{ }315^{\circ}\text{ }(or)\text{ }\frac{7\pi}{4} \end{gathered}[/tex]Final answer:
The values are,
[tex]\theta=\frac{5\pi}{4},\frac{7\pi}{4}[/tex]