We have: #2 (cos(theta))^(2) - cos(theta) - 1 = 0#; #[0, 2 pi]#
This equation is in the form of a quadratic equation.
Let's factorise it using the "middle-term break":
#Rightarrow 2 (cos(theta))^(2) - 2 cos(theta) + cos(theta) - 1 = 0#
#Rightarrow 2 cos(theta)(cos(theta) - 1) + 1 (cos(theta) - 1) = 0#
#Rightarrow (cos(theta) - 1)(2 cos(theta) + 1) = 0#
Using the null factor law:
#Rightarrow cos(theta) - 1 = 0#
#Rightarrow cos(theta) = 1#
#Rightarrow theta = 0, 0 + 2 pi#
#Rightarrow theta = 0, 2 pi#
#or#
#Rightarrow 2 cos(theta) + 1 = 0#
#Rightarrow cos(theta) = - frac(1)(2)#
#Rightarrow theta = pi - frac(pi)(3), pi + frac(pi)(3)#
#Rightarrow theta = frac(2 pi)(3), frac(4 pi)(3)#
Therefore, the solutions are #theta = 0, frac(2 pi)(3), frac(4 pi)(3), 2 pi#.
