how do i solve for x given sinxtan3x = 0 ??? ALSO the range is -2pi < x < 0?

Question

even if i could get the answer..i dont know how to make sure its in the range. i never get negative answers..??

Answer 1

Given:
Sin(x)Tan3(x) = 0
For the equation to be balanced, Sinx or Tan3x, or both must be equal to zero.
The values for Sinx to be zero over the range -2π < x < 0 is -π.
The values for Tan3x to be zero over the range -2π < x < 0 is -π.

If you want a value of x over the range other them zero, then just put the value into the formula, example:
Find value when x = -1∙5.
Sin(x)Tan3(x) =
Sin(-1∙5)Tan3(-1∙5) =
(-0∙997 494 968..)Tan(-4∙5) =
(-0∙997 494 968..)(-4∙637 332 2054...) =
= 4∙625 715 475.. Rads.

Answer 2

From sinx = 0, you have x = -pi

From tan3x = 0, you have 3x = -npi.

Therefore
x = -pi/3, -2pi/3, -pi, -4pi/3 and -5pi/3