Determine solutions for:

tanxcos^2x - tanx = 0

in the interval[-2pi, 2pi]

tanx(cos^2x-1)=-tanxsin^2x

for that to be zero, then sinx is zero, tan x is zero. Where does that occur?

To determine the solutions of the equation tan(x)cos^2(x) - tan(x) = 0 in the interval [-2π, 2π], we can follow these steps:

Step 1: Factor out the common term of tan(x) from the equation.
tan(x)(cos^2(x) - 1) = 0

Step 2: Simplify the expression cos^2(x) - 1.
Recall that cos^2(x) - 1 = sin^2(x). Replace cos^2(x) - 1 with sin^2(x) to rewrite the equation as:
tan(x)(sin^2(x)) = 0

Step 3: Apply the zero-product property.
According to the zero-product property, when the product of two factors is equal to zero, one or both of the factors must be equal to zero. Therefore, we can set each factor to zero and solve for x separately.

Factor 1: tan(x) = 0
To find the solutions for tan(x) = 0, we need to determine when the tangent function is equal to zero. The tangent function is equal to zero at x = 0 and x = π.

Factor 2: sin^2(x) = 0
To determine when sin^2(x) = 0, we solve for x such that sin(x) = 0. The sine function is equal to zero at x = 0, π, 2π, -π, and -2π.

Therefore, the solutions for the given equation in the interval [-2π, 2π] are x = 0, π, 2π, -π, and -2π.