Solve the equation on the interval [0, 2π).

tan2x sin x = tan2x

would it be 0, pi/2 ?

well, you must have

tan2x = 0
or sinx = 1

so, you are almost right. tan2x=0 if 2x = 0 or π or 2π or 3π
so, x = 0, π/2, π, 3π/2
sinx = 1 if x = π/2, so that does not add any solutions to the list.

To solve the equation tan2x sin x = tan2x on the interval [0, 2π), follow these steps:

Step 1: Start by subtracting tan2x from both sides of the equation to isolate the sine term:
tan2x sin x - tan2x = 0

Step 2: Factor out the common factor of tan2x from the left side of the equation:
tan2x (sin x - 1) = 0

Step 3: Set each factor equal to zero and solve for x separately.
First factor:
tan2x = 0

Since tan2x equals zero when the tangent function is zero, the solutions for this factor are:
2x = 0 or 2x = π

Simplifying each equation:
For 2x = 0:
x = 0

For 2x = π:
x = π/2

Second factor:
sin x - 1 = 0

Solving for x:
sin x = 1

The solution for this factor is:
x = π/2

Step 4: Combine all the solutions obtained in steps 3.
x = 0, π/2

Thus, the solutions to the equation tan2x sin x = tan2x on the interval [0, 2π) are 0 and π/2.

To solve the equation tan^2(x) sin(x) = tan^2(x) on the interval [0, 2π), you need to follow these steps:

Step 1: Divide both sides of the equation by tan^2(x) to eliminate the common factor:
sin(x) = 1

Step 2: Now, you have the equation sin(x) = 1. To find the solutions in the interval [0, 2π), you can use the unit circle or a trigonometric identity.

Step 3: On the unit circle, sin(x) = 1 occurs at two points: π/2 and 5π/2.

Step 4: Since you are working within the interval [0, 2π), you should only consider π/2 as a valid solution.

Therefore, the solution to the equation tan^2(x) sin(x) = tan^2(x) on the interval [0, 2π) is x = π/2.