Question

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

tan2x sin x = tan2x

A. 0, (pi)/2

B. (pi)/2 , 2π

C. (pi)/2 , π

D. 0, π

Answer 1

@jim_thompson5910

Answer 2

hint:

tan2x sin x = tan2x

tan2x sin x - tan2x = 0

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

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

Answer 3

i dont understand step 3.. @jim_thompson5910

Answer 4

I factored out tan(2x) using the distributive property

Answer 5

Notice if you distribute tan(2x) back in, you'll get step 2 again

Answer 6

ohh. ok! so 0 is one of the choices. how do I figure out which one is the second choice?

Answer 7

Solve sin(x) - 1 = 0 for x and tell me what you get

Answer 8

pi/2!

Answer 9

so its A?

Answer 10

you got it

Answer 11

thanks!

Answer 12

np