Find all 0 ≤ x < 2π such that 2 sin(2x) = 1.

Answer 1

2sin(2x) = 1

sin(2x) = 1/2

2x = 30 degrees or 150 degrees, or pi/6 or 5pi/6

x = 15 or 75 degrees, = pi/12 or 5pi/12

2x also = 360+30 or 360+150 = 390 or 510 degrees

x then = also 195 or 255 degrees =13pi/12 or 17pi/12

x = 15, 75, 195 or 255 degrees

= pi/12, 5pi/12, 13pi/12, 17pi/12

2sin2(15) = 2sin30 = 2(1/2) = 1

2sin2(75) = 2sin150 = 2(1/2) = 1

2sin2(195) = 2sin(390) = 2sin30=1

2sin2(255) = 2sin510 = 2sin150= 1

Answer 2

The equation is sin(2x) = 1/2. There are two angles θ in [0, 2π) with sin(θ) = 1/2, namely θ = π/6, 5π/6, so we get 2x = π/6 + 2πk or 2x = 5π/6 + 2πk, where k is any integer. This gives x = π/12 + πk or x = 5π/12 + πk for the general solution of the equation. Now pick out those values in [0, 2π]: x = π/12, 5π/12, 13π/12, 17π/12.