Given:
The function is:
[tex]2\cos2x-1=0[/tex]
Find-:
The value of "x"
Explanation-:
The value of x is:
[tex]\begin{gathered} 2\cos2x-1=0 \\ \\ 2\cos2x=1 \\ \\ \cos2x=\frac{1}{2} \\ \end{gathered}[/tex]
Solve for x is:
[tex]\begin{gathered} \cos2x=\frac{1}{2} \\ \\ 2x=\cos^{-1}(\frac{1}{2}) \\ \\ 2x=\frac{\pi}{3}+2\pi n\text{ and }2x=\frac{5\pi}{3}+2\pi n \end{gathered}[/tex]
The value of "x" is:
[tex]\begin{gathered} 2x=\frac{\pi}{3}+2\pi n \\ \\ x=\frac{\pi}{2\times3}+\frac{2\pi n}{2} \\ \\ x=\frac{\pi}{6}+\pi n \end{gathered}[/tex]
Another value of "x" is:
[tex]\begin{gathered} 2x=\frac{5\pi}{3}+2\pi n \\ \\ x=\frac{5\pi}{3\times2}+\frac{2\pi n}{2} \\ \\ x=\frac{5\pi}{6}+\pi n \end{gathered}[/tex]
So, the answer is:
[tex]x=\frac{\pi}{6}+\pi n,\frac{5\pi}{6}+\pi n[/tex]
