solve the equation tan^2 (theta) + tan (theta)=0 for the interval 0 is less than or equal to (theta) is less than 2 pie.

To solve the equation tan^2(theta) + tan(theta) = 0 for the interval 0 ≤ θ < 2π, we can use factoring.

First, let's rewrite the equation in terms of tan(theta):

tan^2(theta) + tan(theta) = 0

Using factoring, we can factor out a common factor of tan(theta):

tan(theta) * (tan(theta) + 1) = 0

Now we have two factors: tan(theta) = 0 and tan(theta) + 1 = 0

1. For tan(theta) = 0, we know that tan(theta) is equal to zero when theta is an angle where the tangent function equals zero. In the interval 0 ≤ θ < 2π, these angles are θ = 0 and θ = π.

2. For tan(theta) + 1 = 0, we can solve for theta:

tan(theta) = -1

To find the values of theta where the tangent function equals -1, we can use the inverse tangent (arctan) function. Taking the arctan of both sides of the equation, we get:

θ = arctan(-1)

The principal value of arctan(-1) is -π/4. However, we need to find all the solutions in the given interval.

Adding π to the principal value (-π/4 + π), we get 3π/4.

Now we have three solutions: θ = 0, θ = π, and θ = 3π/4.

Therefore, the solutions to the equation tan^2(theta) + tan(theta) = 0 for the interval 0 ≤ θ < 2π are θ = 0, θ = π, and θ = 3π/4.