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