Given: A function
![f(x)=(1)/(2)\cos2x-1](https://img.qammunity.org/2023/formulas/mathematics/college/7he4cxh8kfm63i9baaw7y0xnu9tylgz4x5.png)
Required: Zeroes of the function in the interval [0,2π]
Step-by-step explanation:
For zero of the function f(x)
![\begin{gathered} f(x)=0 \\ (1)/(2)\cos2x-1=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kmaj6st2kkuddwx28fggmdqs3937guzkcg.png)
Now,
![\begin{gathered} (1)/(2)\cos2x=1 \\ \cos2x=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z2bf8ts7ubushcjen3s5q4ubo7w01dhpyf.png)
Now, cos function alsways lies between [-1,1].
It is not equal to 2 in the interval [0,2π].
Hence, the zeroes does not exist in this interval.
Final Answer: The correct answer is option 1.