solve w/o calculator, give answer in exact radians, range: 0-2pi

Question

2cos(2x)+1=0

solve this equation w/o using a calculator please

Answer 1

Hi Sibel O.,

2cos(2x)+1=0

2cos(2x) = -1, subtract 1 from both sides

cos(2x) = -1/2, divide both sides by 2

2*cos²(x) - 1 = -1/2, from the identity cos(2u) = 2*cos²(u) - 1

2*cos²(x) = 1/2, add 1 to both sides

cos²(x) = 1/4, divide both sides by 2

cos(x) = ± 1/2, square root both sides

So for 0 < x < 2π, cos(x) = ±1/2 when x = π/3, 2π/3, 4π/3, 5π/3.

Lets check:

2*cos[2*(π/3)] + 1 = 0

2*(-1/2) + 1 = 0, true for π/3

2*cos[2*(2π/3)] + 1 = 0

2*cos(4π/3) + 1 = 0

2*(-1/2) + 1 = 0, true for 2π/3

2*cos[2*(4π/3)] + 1 = 0

2*cos(8π/3) + 1 = 0

2*(-1/2) + 1 = 0, true for 4π/3

2*cos[2*(5π/3)] + 1 = 0

2*cos(10π/3) + 1 = 0

2*(-1/2) + 1 = 0, true for 5π/3

I hope this helps, Joe.

Answer 2

2cos(2x)+1=0

2cos(2x)=-1

cos(2x)=-1/2

cos=-1/2 when 2x=2∏/3 and when 2x=4∏/3

therefore x=∏/3 and x=2∏/3