Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Is there a number whose cosine is itself? Let , with in radians. Find and show that changes sign on . Then halve the interval three times, finding along the way, and give the width of the interval that remains.
If is continuous on and and have opposite signs, then for some in . Halving the interval and keeping the half with the sign change traps the root.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The sign change. and . A root lies in .
Step 2. First halving. , so the root is in .
Step 3. Second halving. , so the root is in .
Step 4. Third halving. , so the root is in . Its width is .
Answer: , , and the width is .
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