please solve this ......2cos^2x + cos x - 1..........could you please elaborate in the answer on how you got?

Question

THANKS!!!

Answer 1

(2cosx+1)(cosx−1)=0,
so solve each for when either equals 0,
cosx−1=0 when x=360n, and 2cosx+1=0 when x=360n±120
'n' being any positive/negative integer

Answer 2

You could also do the substitution y=cos(x) so that it simplifies to something you are probably more familiar with...

2y^2+y-1....... solve with quadratic equation and then solve back for x using y=cos(x) after youve solved for y.

also, from above, assuming this whole thing equals zero

(2*cos(x)- 1)*(cos(x)+1)=0

then the solutions are cos(x) = 1/2 and cos(x) = -1

note that there are actually an infinite number of solutions

cos(x) = 1/2 => x = pi/6 + 2*n*pi and x = -pi/6 + 2*n*pi for n=0,1,2,3....infinity

same thing for cos(x) = -1

Answer 3

2cos^2x + cos x - 1
= (2Cos x-1) (Cos x+1)
this is an expression and not an equation...
so there wont be a solution

Answer 4

Factor it: (2cosx + 1)(cosx - 1) = 0.

2cosx + 1 = 0 --> cos(x)=0.5, so x=arccos(0.5) = (pi/3) (considering only a simple answer).

cos(x) - 1 = 0 --> cos(x) = 1 --> x = 0.

There are infinitely many answers due to the periodicity of cos(x).

Answer 5

2cos^2x+cosx-1=0
2cos^2x+2cosx-cosx-1
=2cosx(cosx+1)-1(cosx+1)=0
(cosx+1)(2cosx-1)=0
cosx=1,-1/2
x=0,120,240deg
=0,2pi/3, 4pi/3.answer