You should also use the following alternative method such that:

`6 cos^2 theta = 1 - cos theta` 

You need to substitute `2sin^2(theta/2)`  for `1 - sin theta`  such that:

`6 cos^2 theta = 2sin^2(theta/2)`

You should write `cos^2 theta`  such that:

`cos^2 theta = cos^2 2(theta/2)`

Substituting `1 - 2 sin^2(theta/2)`  for `cos^2 2(theta/2)`  yields:

`6cos^2 2(theta/2) = 2sin^2(theta/2) => 6(1 - 2 sin^2(theta/2)) = 2sin^2(theta/2)`

Reducing by 2 both sides yields:

`3(1 - 2 sin^2(theta/2)) = sin^2(theta/2)`

Opening the brackets yields:

`3 - 6sin^2(theta/2) = sin^2(theta/2)`

`3 = 7sin^2(theta/2) => sin^2(theta/2) = 3/7`

`sin (theta/2) = +-sqrt(3/7) ` `theta/2 = arcsin(+-sqrt(3/7)) => theta = 2sin(+-sqrt(3/7)) => theta = +-2sin(sqrt(3/7)) `

You may use graphical method to find the solutions to the give equation, such that:

Hence, evaluating the solutions to the given equation yields `theta = +-2sin(sqrt(3/7)).`

First use substitution `cos theta=t.` Now equation becomes:

`6t^2-t-1=0` Now we solve quadratic equation

`t_(1,2)=(-(-1)pm sqrt(1-4cdot6cdot(-1)))/(2cdot6)=(1pm5)/(12)`

`t_1=-1/3`, `t_2=1/2`

So we have

`cos theta=-1/3=> theta=pm arccos(-1/3)+2k pi, k in ZZ` and

`cos theta=1/2=> theta={pi/3+2k pi,(2pi)/3+2k pi; k in ZZ}.`

So there are 4 solutions for `0leq theta leq 2pi` and they are:

`theta_1=arccos(1/3)`, `theta_2=-arccos(1/3)+2pi`, `theta_3=pi/3` and `theta_4=(2pi)/3`.