Remember the following trigonometric identities:
[tex]\begin{gathered} \sin (x)=\cos (x-\frac{\pi}{2}) \\ \sin (x+2\pi k)=\sin (x)\text{ for all integer values of k} \\ \cos (x)=a\Rightarrow x=\pm\cos ^{-1}(a) \\ \sin (-x)=-\sin (x) \\ -\cos (x)=\cos (x-\pi) \end{gathered}[/tex]Using these identities, search for an angle x such that:
[tex]\sin (x)=-\frac{1}{2}[/tex]Write the sine of x in terms of the cosine:
[tex]\Rightarrow\cos (x-\frac{\pi}{2})=-\frac{1}{2}[/tex]Multiply both sides by -1:
[tex]\Rightarrow-\cos (x-\frac{\pi}{2})=\frac{1}{2}[/tex]Since -cos(x)=cos(x-π), then:
[tex]\begin{gathered} \Rightarrow\cos (x-\frac{\pi}{2}-\pi)=\frac{1}{2} \\ \Rightarrow cos(x-\frac{3}{2}\pi)=\frac{1}{2} \end{gathered}[/tex]Use the inverse cosine function to isolate x-3π/2:
[tex]\Rightarrow x-\frac{3}{2}\pi=\pm\cos ^{-1}(\frac{1}{2})[/tex]Since cos(π/3) = 1/2, then cos^-1(1/2) = π/3:
[tex]\begin{gathered} \Rightarrow x-\frac{3}{2}\pi=\pm\frac{\pi}{3} \\ \Rightarrow x=\frac{3}{2}\pi\pm\frac{1}{3}\pi \\ \Rightarrow x=(\frac{3}{2}\pm\frac{1}{3})\pi \\ \Rightarrow x_1=\frac{11}{6}\pi,x_2=\frac{7}{6}\pi \end{gathered}[/tex]To find other values of x which make sin(x) equal to -1/2, we can add integer multiples of 2π to the values x₁ and x₂. In this case, since all options are less than 2π, that is not necessary.
Therefore, the angles whose sine is -1/2 are:
[tex]\begin{gathered} \frac{7}{6}\pi \\ \frac{11}{6}\pi \end{gathered}[/tex]