How would you solve this problem? Please include steps for me!

Solve for θ in the interval 0 ≤ θ ≤ 2π: cos(θ )sin(θ ) − 1/ 2cos(θ ) = 0

If you mean

cos(θ ) sin(θ ) − (1/ 2) cos(θ ) = 0
that is
cos(θ ) [ sin(θ ) - 0.5 ]
this is true if
cos(θ ) = 0
or if
sin(θ ) = 0.5

cos is zero at π/2 and at 3π/2
sin is zero at 0 and at π ( and at 2π which is zero again of course)

sorry sin = 1/2 at π/6 and at 5 π/6

To solve the given equation for θ in the interval 0 ≤ θ ≤ 2π, we can follow these steps:

Step 1: Simplify the equation
Start by factoring out the common term cos(θ):
cos(θ)(sin(θ) - 1/2) = 0

Step 2: Set each factor equal to zero
Since the product of two factors is zero, we can set each factor equal to zero and solve for θ individually.

First, set cos(θ) = 0:
cos(θ) = 0

To find the values of θ that satisfy this equation, we need to recall the unit circle. The unit circle represents the values of sine and cosine for different angles. The cosine is equal to zero at π/2 and 3π/2 on the unit circle.

So, θ = π/2 and θ = 3π/2 are the values that satisfy cos(θ) = 0 in the given interval.

Next, set sin(θ) - 1/2 = 0:
sin(θ) - 1/2 = 0

To solve this equation, we need to isolate sin(θ):
sin(θ) = 1/2

Again, referring to the unit circle, we can see that θ = π/6 and θ = 5π/6 are the values that satisfy sin(θ) = 1/2 in the given interval.

Step 3: Combine all solutions
Now, we have the following values for θ: θ = π/2, θ = 3π/2, θ = π/6, and θ = 5π/6.

Therefore, the solutions for θ in the interval 0 ≤ θ ≤ 2π are: θ = π/2, θ = 3π/2, θ = π/6, and θ = 5π/6.