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

Given

tan (x) ²·sin (x) = tan (x) ²

Find

x on the interval [0, 2π)

Solution

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}