Solve the equation 4sin(x-π)cos(x-π) =√3 for 0≤x≤2π

Step-by-step explanation:

4sin(x-π)cos(x-π) =√3

let (x-π) = (y)

2(2sin(y)cos(y)) = √3

sin(2y) = ½√3

  2y = arcsin(½√3)

  2y = ±π/3 + 2πn, ±2π/3 + 2πn

    y = ±π/6 + πn, ±π/3 + πn

x - π = ±π/6 + πn, ±π/3 + πn

    x = ±π/6 + π(n + 1), ±π/3 + π(n + 1)

As n = any whole number

    x = ±π/6 + π(n), ±π/3 + π(1)

so in the range 0 ≤ x ≤ 2π

x = + π/6 + 0π = π/6

x = + π/6 + 1π = 7π/6

x = - π/6 + 1π = 5π/6

x = - π/6 + 2π = 11π/6

x = + π/3 + 0π = π/3

x = + π/3 + 1π = 4π/3

x = - π/3 + 1π = 2π/3

x = - π/3 + 2π = 5π/3

in increasing magnitude

x = π/6, π/3, 2π/3, 5π/6 7π/6, 4π/3, 5π/3, 11π/6

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah