How do you solve 2cos²(x)-cos(x)=0

Step-by-step explanation:

Start by factoring the left side of 2cos²(x)-cos(x)=0:

cos x [ 2cos x - 1 ] = 0.  

Then cos x = 0 and 2cos x - 1  = 0.

Restricting ourselves to positive angles between 0 and 2π, we get from cos x = 0 the angles x = π/2 and x = 3π/2.

Rewriting 2cos x - 1  = 0, we get

2cos x = 1, or cos x = 1/2.  One solution is x = π/3, in the first quadrant; the other is x = 5π/3, in the fourth quadrant.

i would pic false cause it goes down towards negative

step-by-step explanation: