Answer: To solve the equation (cosx + 1)(2cos^2 x - 3cosx - 2) = 0, we'll first find the values of x that make each factor equal to zero and then find the corresponding solutions for x in the given interval [0, 2π).
Step 1: Set each factor equal to zero and solve for x:
cosx + 1 = 0
Subtract 1 from both sides:
cosx = -1
2cos^2 x - 3cosx - 2 = 0
This is a quadratic equation. Let's factor it:
2cos^2 x - 3cosx - 2 = (2cosx + 1)(cosx - 2)
Now, set each factor equal to zero:
2cosx + 1 = 0
cosx = -1/2
cosx - 2 = 0
cosx = 2 (Note: Cosine values are always between -1 and 1, so there is no solution for this equation.)
Step 2: Find the solutions for x in the interval [0, 2π):
For cosx = -1, the solutions in the given interval are π (180 degrees) and 3π (540 degrees).
For cosx = -1/2, the solutions in the given interval are π/3 (60 degrees) and 5π/3 (300 degrees).
So, the solutions for x in the interval [0, 2π) are:
x = π, π/3, and 5π/3.