We are given the following equation:
![\cos 2x=(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/veqpvmn4h24i0xap0lmrmsiri7wf3kszu3.png)
To solve for "x" we will take arccos to both sides:
![2x=\text{arccos}((1)/(2))](https://img.qammunity.org/2023/formulas/mathematics/high-school/sotibkwa9tuu0hnoviozqcsidnnrsrwywx.png)
Solving the operations:
![2x=(\pi)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/dkrg9mwonv6oi78mciede1hx08hsy3ztd4.png)
This is for the first quadrant. Dividing both sides by 2:
![x=(\pi)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/cwrlj3e9zwrz2mtenu0sx1ppidfoefrdgb.png)
For the second quadrant we have:
![2x=(5\pi)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/lhk6moenrb5jdqo7bt9vnh1n4duersmkyc.png)
Dividing both sides by 2:
![x=(5\pi)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/55hilackj6kkr7erqlaarrax01rmdvw10e.png)
For the third quadrant we have:
![2x=(7\pi)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/ugc5qhos6202c4ldoh2nqgy9c0exg3rkbt.png)
Dividing by 2:
![x=(7\pi)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/1bo2ai38nax1smgsr8mcmnb3fpmwb06gkx.png)
For the fourth quadrant:
![2x=(11\pi)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/cfdyh0ezuj8r1ld6m67hof9ndldcdngwd6.png)
Dividing by 2:
![x=(11\pi)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/dfqb6esffihpz5usxlmqqjl3aqoau3t2zt.png)
Therefore, the values of "x" are:
![x=(\pi)/(6),(5\pi)/(6),(7\pi)/(6),(11\pi)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/eijce7zmlg8ptfjhkfvk6y8kfoc7939r83.png)