find all the values of X between 0 and 2π such that cos(3x-π/2) = -√3/2
cos(3x- ƒÎ/2) = sin(3x)
(Check your trig identities)
3x = sin^-1(-�ã3/2)
= 4 pi/3, 5 pi/3, 10 pi/3, 11 pi/3 etc
x = 4 pi/9 etc
To find all the values of x between 0 and 2π that satisfy the equation cos(3x - π/2) = -√3/2, we can follow these steps:
Step 1: Identify the corresponding angle with a cosine value of -√3/2. One such angle is π/6 or 30 degrees.
Step 2: Use the identity cos(θ) = cos(2π - θ) to find all possible angles that satisfy the equation. This identity states that the cosine function is symmetric around the x-axis. Therefore, if θ is a solution, then (2π - θ) is also a solution.
Using this identity, we have:
3x - π/2 = π/6 ......................... (1) [Using the initial angle π/6]
3x - π/2 = 2π - π/6 ................. (2) [Using the identity]
Step 3: Solve each equation separately to find the values of x.
Equation (1):
3x - π/2 = π/6
3x = π/6 + π/2
3x = π/6 + 3π/6
3x = 4π/6
x = (4π/6) / 3
x = 4π/18
x = 2π/9
Equation (2):
3x - π/2 = 2π - π/6
3x = 2π - π/6 + π/2
3x = 2π + 2π/6 - π/6
3x = 12π/6
x = (12π/6) / 3
x = 4π/6
x = 2π/3
Therefore, the values of x between 0 and 2π that satisfy the equation cos(3x - π/2) = -√3/2 are x = 2π/9 and x = 2π/3.