Solve the following equation on the interval [0, 2π). tan2x sin x = tan2x

tan(2x)·sin(x) = tan(2x)

x on the interval [0, 2π)

Subtract the right side and factor. Invoke the zero-product rule.

... tan(2x)sin(x) -tan(2x) = 0

... tan(2x)(sin(x) -1) = 0

... tan(2x) = 0

... 2x = arctan(0) = nπ

... x = (n/2)π . . . . n = {0, 1, 2, 3}

And when the other factor is zero, we have

... sin(x) -1 = 0

... sin(x) = 1

... x = arcsin(1) = π/2

So, we have

... x ∈ {0, π/2, π, 3π/2}

step-by-step explanation: