To find the x-intercepts if the function we equate to zero and solve for x:
[tex]\begin{gathered} 2\sin (2x)-1=0 \\ 2\sin (2x)=1 \\ \sin (2x)=\frac{1}{2} \end{gathered}[/tex]
Now we need to remember that the sine function is equal to 1/2 if:
[tex]\begin{gathered} x=\frac{1}{6}(12\pi n+\pi) \\ \text{ or } \\ x=\frac{1}{6}(12\pi n+5\pi) \end{gathered}[/tex]
Then we have that:
[tex]\begin{gathered} 2x=\frac{1}{6}(12\pi n+\pi) \\ x=\frac{1}{12}(12\pi n+\pi) \\ \text{ or } \\ 2x=\frac{1}{6}(12\pi n+5\pi) \\ x=\frac{1}{12}(12\pi n+5\pi) \end{gathered}[/tex]
Now we need to give values to n to find the vaues of x in the interval. Using the first expression we have:
[tex]\begin{gathered} x=\frac{1}{12}(12\pi\cdot0+\pi)=\frac{1}{12}\pi \\ x=\frac{1}{12}(12\pi\cdot1+\pi)=\frac{13}{12}\pi \end{gathered}[/tex]
using the second expression we have:
[tex]\begin{gathered} x=\frac{1}{12}(12\pi\cdot0+5\pi)=\frac{5}{12}\pi \\ x=\frac{1}{12}(12\pi\cdot1+5\pi)=\frac{17}{12}\pi \end{gathered}[/tex]
Therefore the x-intercepts in the interval given is:
[tex]\lbrace\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\rbrace[/tex]
