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

tan(x)²·sin(x) = tan(x)²

x on the interval [0, 2π)

Subtract the right side and factor. Then make use of the zero-product rule.

... tan(x)²·sin(x) -tan(x)² = 0

... tan(x)²·(sin(x) -1) = 0

This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:

... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)

Then our equation becomes

... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0

... -sin(x)²/(1 +sin(x)) = 0

Now, we know the only solutions are found where sin(x) = 0, at ...

... x ∈ {0, π}

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