gloriallanas
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# Solve 2cos^2x + cosx − 1 = 0 for x over the interval [0, 2 pi ).

2cos²x + cosx − 1 = 0 (cosx+1)(2cosx-1)=0 so cosx+1=0 or 2cosx-1=0 1) cosx+1=0 cosx=-1 x=π+2πk, k∈Z 2) 2cosx-1=0 2cosx=1 cosx=0.5 x=+-$\frac{ \pi }{6}+2 \pi k$, k∈Z find value for the interval [0, 2 pi ) {$\frac{ \pi }{6}, \pi , - \frac{ \pi }{6}+2 \pi k$} Answer: $\frac{ \pi }{6}, \pi , \frac{ 11\pi }{6}$
$2cos^{2}(x) + cos(x)-1 = 0$ 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: $x = \boxed{0, \frac{ \pi }{3}, \frac{5 \pi }{3}}$