`y = cos(2x)` and `y= -1/2`

Let's find the general solution in `-piltxltpi ` .

`cos(2x) = -1/2`

The primary solution for `(2x)` is,

`2x = cos^(-1)(-1/2)`

`(2x) = (2pi)/3`

The general solution for `cos(2x)` is given by,

`2x = 2npi+-(2pi)/3`

`x = npi+-pi/3` where `n in Z` .

For n= -1,

`x = -pi+-pi/3 `

The solution in the given range is `-pi+pi/3 = (-2pi)/3`

For n = 0,

`x = +-pi/3,`

The solutions in teh given range are, `pi/3` and `(-pi)/3` .

For n= 1,

`x = pi+-pi/3`

The solution in teh given range is,

x = pi-pi/3 = (2pi)/3

**Therefore the solutions for x in the given range are,**

`x = (-2pi)/3` ,` x = (-pi)/3` ,` x =pi/3` and `x = (2pi)/3`