Solve the following equations: cos(2x) = -1/2 , 0 <= x <= 2π

a) cos(2x) = -1/2 , 0 <= x <= 2π

b) 2sin(3x) - √3 = 0 , 0 <= x <= 2π

a) Set both sides by arccos and divide both sides by 2 to get:

x = arccos(-½)/2
x = π/3, 4π/3, 2π/3, 5π/3

b) 2sin(3x) - √(3) = 0

Rearrange the terms...

sin(3x) = √(3)/2
x = arcsin(√(3)/2)/3
x = π/9, 2π/9, 7π/9

I hope this helps!

It x is on the interval [0, 2pi), then 2x is on the interval [0, 4pi) and 3x is on [0, 6pi).

a) Cos 2x = - 1/2 when 2x = 2pi/3 and 4pi/3; on the interval [0, 4pi) these are coterminal with 8pi/3 and 10pi/3. So x = pi/3, 2pi/3, 4pi/3, and 5pi/3.

b)sin(3x) = sqrt(3)/2 when 3x = pi/3 and 2pi/3; on the interval [0, 6pi), these are coterminal with 7pi/3 and 8pi/3, 13pi/3 and 14pi/3. So x = pi/6, pi/3, 7pi/6, 4pi/3, 13pi/6, and 7pi/3.