Question

PLEASE !!!! for a medal Solve 2cos^2x + cosx − 1 = 0 for x over the interval [0, 2 π ).
a.π and π/3
b.π, π/3, and 5π/3
c.1 and 2π/3
d.1, 2π/3, and 4π/3
e.1, π/3, and 5π/3

Answer 1

\[2\cos ^2x+2\cos x-\cos x-1=0\]
make factors and find value of cos x and then x

Answer 2

idk how

Answer 3

and i dont have time

Answer 4

take cos x common from first two and -1 from last two

Answer 5

cos^2x + cosx = 0?

Answer 6

\[2\cos x \left( \cos x+1 \right)-1\left( \cos x+1 \right)=0\]
\[\left( \cos x+1 \right)\left( 2\cos x-1 \right)=0\]
equate each factor to 0 and find value of x.
will you show me ?

Answer 7

either cos x+1=0,
cos x=-1
can you tell at what angle value of cosx =-1

Answer 8

idk

Answer 9

x=π±2πn,π/3±2πn,5π/3±2πn

Answer 10

i got that thing for first thing you asked

Answer 11

\[\cos x=-1=\cos \pi \in 0 \le x \le 2 \pi,x=\pi\]

Answer 12

\[2 \cos x=1,\cos x=\frac{ 1 }{ 2 }=\cos \frac{ \pi }{ 3 },\cos \left( 2\pi-\frac{ \pi }{ 3 } \right)\]
x=?

Answer 13

cos(5x/3)

Answer 14

wait i made a mistake

Answer 15

1/2