Final answer:
The domain of 1/cos(2x) is all real numbers except for values where cos(2x) equals zero, which occurs at x = (2k+1)\(¼\pi\), thus the correct domain is Option C: x in k\(π\) + \(¼\pi\) .
Step-by-step explanation:
The domain of a function refers to all the possible input values (x) that the function can accept without resulting in an undefined or nonreal number. For the function 1/cos(2x), the denominator, cos(2x), cannot be equal to zero because division by zero is undefined. The cosine function equals zero whenever its argument is an odd multiple of \(½\pi\).
So, we must exclude the values of x where cos(2x)=0. This occurs when 2x is an odd multiple of \(½\pi\), which can be expressed as 2x = (2k+1)\(½\pi\) for any integer k. Solving for x gives us x = (2k+1)\(¼\pi\).
Therefore, the domain of 1/cos(2x) is all real numbers excluding those where 2x is an odd multiple of \(½\pi\). The correct answer is Option C: x in k\(π\) + \(¼\pi\) , meaning the domain of 1/cos(2x) is all real numbers except those that can be expressed as k\(π\) + \(¼\pi\) where k is any integer.