To solve this inequality, we can first rewrite the given expression using trigonometric identities.
Recall that sin(2θ) = 2sinθcosθ. Therefore, we have:
cos(3x-π/3)sin(3x-π/3) = 1/2 * sin(6x-2π/3)
Now we need to find the solutions for sin(6x-2π/3)≥1.
Since sinθ = 1 when θ = π/2 + 2kπ, where k is an integer, we solve for:
6x-2π/3 = π/2 + 2kπ
Solving for x:
6x = 2π/3 + π/2 + 2kπ6x = 3π/6 + 2π/6 + 12kπ/6x = π/6 + π/6 + 2kπ/6x = π/3 + kπ/3
Therefore, the solution to the inequality is x = π/3 + kπ/3, where k is an integer.
To solve this inequality, we can first rewrite the given expression using trigonometric identities.
Recall that sin(2θ) = 2sinθcosθ. Therefore, we have:
cos(3x-π/3)sin(3x-π/3) = 1/2 * sin(6x-2π/3)
Now we need to find the solutions for sin(6x-2π/3)≥1.
Since sinθ = 1 when θ = π/2 + 2kπ, where k is an integer, we solve for:
6x-2π/3 = π/2 + 2kπ
Solving for x:
6x = 2π/3 + π/2 + 2kπ
6x = 3π/6 + 2π/6 + 12kπ/6
x = π/6 + π/6 + 2kπ/6
x = π/3 + kπ/3
Therefore, the solution to the inequality is x = π/3 + kπ/3, where k is an integer.