Answer:
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.