[tex]2cos^{2}(x) + cos(x)-1 = 0[/tex]
This could also be written as, where [tex]a = cos(x)[/tex]
[tex]2a^{2} + a - 1 = 0[/tex]
This would factorize to give:
[tex](2a-1)(a+1)=0[/tex]
So we can factorize our original expression:
[tex]2cos^{2}(x) + cos(x)-1 = 0 \\ \\ (2cosx - 1)(cosx+1) = 0 [/tex]
We can then solve for [tex]x[/tex] as we would with a normal quadratic:
[tex]2cosx -1 =0 \\ \\ cosx = \frac{1}{2} \\ \\ x = cos^{-1}( \frac{1}{2} ) \\ \\ x = \frac{ \pi }{3}, \frac{5 \pi }{3} [/tex]
And also:
[tex]cos(x)+1 = 0 \\ \\ cos(x)= -1 \\ \\ x = cos^{-1}(1) \\ \\ x = 0, 2 \pi [/tex]
So our values for [tex]x[/tex] are:
[tex]x =0, \frac{ \pi }{3}, \frac{5 \pi }{3}, 2 \pi[/tex]
As: [tex]0 \leq x\ \textless \ 2 \pi [/tex]
Our final solutions for [tex]x[/tex] are:
[tex]x = \boxed{0, \frac{ \pi }{3}, \frac{5 \pi }{3}}[/tex]
