Question

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

Answer 1

$2cos^{2}(x) + cos(x)-1 = 0$

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:

$x = \boxed{0, \frac{ \pi }{3}, \frac{5 \pi }{3}}$

Answer 2

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: $\frac{ \pi }{6}, \pi , \frac{ 11\pi }{6}$