Find all solutions in the interval [0, 2π). 7 tan3x – 21 tan x = 0?

2 Answers

• Is that tan(3x) or tan(x)^3? Do you see these things: ( ). They’re important

  I’ll do both

  7 * tan(3x) – 21 * tan(x) = 0

  tan(3x) – 3 * tan(x) = 0

  tan(2x + x) – 3 * tan(2x – x) = 0

  (tan(2x) + tan(x)) / (1 – tan(2x) * tan(x)) = 3 * (tan(2x) – tan(x)) / (1 + tan(2x) * tan(x))

  (tan(2x) + tan(x)) * (1 + tan(2x) * tan(x)) = 3 * (tan(2x) – tan(x)) * (1 – tan(2x) * tan(x))

  (tan(2x) + tan(x)) * 1 + (tan(2x) + tan(x)) * (tan(2x) * tan(x)) = 3 * (tan(2x) – tan(x)) * 1 – 3 * (tan(2x) – tan(x)) * tan(2x) * tan(x)

  (tan(2x) * tan(x)) * (tan(2x) + tan(x)) + 3 * tan(2x) * tan(x) * (tan(2x) – tan(x)) = 3 * (tan(2x) – tan(x)) – (tan(2x) + tan(x))

  tan(2x) * tan(x) * (tan(2x) + tan(x) + 3 * tan(2x) – 3 * tan(x)) = 3 * tan(2x) – tan(2x) – 3 * tan(x) – tan(x)

  tan(2x) * tan(x) * (4 * tan(2x) – 2 * tan(x)) = 2 * tan(2x) – 4 * tan(x)

  2 * tan(2x) * tan(x) * (2 * tan(2x) – tan(x)) = 2 * (tan(2x) – 2 * tan(x))

  tan(x) * tan(2x) * (2 * tan(2x) – tan(x)) = tan(2x) – 2 * tan(x)

  tan(x) * tan(2x) * (2 * tan(2x) – tan(x)) = (2 * tan(x) / (1 – tan(x)^2)) – 2 * tan(x)

  tan(x) is a common factor on both sides, so tan(x) = 0 is a possible solution set

  tan(2x) * (2 * tan(2x) – tan(x)) = 2 / (1 – tan(x)^2) – 2

  tan(2x) * (2 * tan(2x) – tan(x)) = (2 – 2 + 2 * tan(x)^2) / (1 – tan(x)^2)

  2 * tan(x) * (2 * tan(2x) – tan(x)) / (1 – tan(x)^2) = 2 * tan(x)^2 / (1 – tan(x)^2)

  Again tan(x) = 0 is a possible solution set and we can also rule out 1 – tan(x)^2 = 0 as a solution set. That is, if tan(x) = +/- 1, then it’s not part of the solution set

  2 * (2 * tan(2x) – tan(x)) = 2 * tan(x)

  2 * tan(2x) – tan(x) = tan(x)

  2 * tan(2x) = 2 * tan(x)

  tan(2x) = tan(x)

  2 * tan(x) / (1 – tan(x)^2) = tan(x)

  Again, tan(x) = 0 is a part of our solution set

  2 / (1 – tan(x)^2) = 1

  2 = 1 – tan(x)^2

  tan(x)^2 = -1

  No solution there.

  tan(x) = 0 is the only possible solution

  x = 0 , pi

  If you meant: 7 * tan(x)^3 – 21 * tan(x) = 0

  7 * tan(x) * (tan(x)^2 – 3) = 0

  tan(x) = 0

  x = 0 , pi

  tan(x)^2 – 3 = 0

  tan(x)^2 = 3

  tan(x) = +/- sqrt(3)

  x = pi/3 , 2pi/3 , 4pi/3 , 5pi/3

  x = 0 , pi/3 , 2pi/3 , pi , 4pi/3 , 5pi/3

• I don`t trust this question !

  I have a sneaking suspicion that it should be shown as :-

  7 tan³x – 21 tan x = 0

  7 tan x [ tan²x – 3 ] = 0

  tan x = 0 , tan x = ± √3

  x = 0 , π , π/3 , 2π/3 , 4π/3 , 5π/3