Finding roots is a range of an equation. Help!!!!

[TheHerbalis]

First of all this is my first post here so sorry if this is in the wrong place. Please let me know if it is.

Im really stuck on this question and have no idea how to tackle it. I would apprieciat if someone could run through how to slove it for me.

Find all the roots in the range − pi < x < pi of the equation


cos(2x) = 1/2 giving answers in radians as multiples of Pi.

Thanks in advance

 

[Ishuda]

What is t when cos(t)=1/2? Then substitute t = 2x. As an example, consider sin(2x) = 1/2. Well sin(t) = 1/2 when
t = \(\displaystyle \pi\)/6 + 2 n \(\displaystyle \pi\)
and
t = 5\(\displaystyle \pi\)/6 + 2 n \(\displaystyle \pi\)
where n is an integer = 0, \(\displaystyle \pm\)1, \(\displaystyle \pm\)2, \(\displaystyle \pm\)3, ...

That translates to
2x = \(\displaystyle \pi\)/6 + 2 n \(\displaystyle \pi\)
and
2x = 5\(\displaystyle \pi\)/6 + 2 n \(\displaystyle \pi\)
or
x = \(\displaystyle \pi\)/12 + n \(\displaystyle \pi\)
and
x = 5\(\displaystyle \pi\)/12 + n \(\displaystyle \pi\)
The x's which satisfy -\(\displaystyle \pi\) < x < \(\displaystyle \pi\) or -1 < x/\(\displaystyle \pi\) < 1
The solution is for n=-1 and n=0 or
x/\(\displaystyle \pi\) = -11/12, -7/12, 1/12, and 5/12