Question

6 tan(2x) − 6 cot(x) = 0 ; [0, 2π)

Answer 1

9514 1404 393

Answer:

  x ∈ {π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6}

Step-by-step explanation:

We can divide by 6 and use some identities to rewrite this as ...

  2tan(x)/(1-tan(x)²) -1/tan(x) = 0

  2tan(x)² -(1 -tan(x)²) = 0 . . . . . . . . . multiply* by tan(x)(1 -tan(x)²)

  3tan(x)² = 1

  tan(x) = ±1/√3   ⇒   x = {π/6, 5π/6, 7π/6, 11π/6}

__

Interestingly, if we use different identities, we can rewrite it as ...

  2cot(x)/(cot(x)² -1) -cot(x) = 0

  cot(x)(3 -cot(x)²) = 0 . . . . . . . multiply by (cot(x)² -1)

In addition to the zeros found above, this equation also has solutions where cot(x) = 0. That is ...

  x = {π/2, 3π/2}

_____

Usually, the first method shown would give all the solutions. Fortunately, we had a graph of the function, so we could see that the first set of solutions found was not complete. (This is one reason I like to use a graphing calculator for guidance on problems like this.)

_____

* In the first solution, we multiplied by the denominator tan(x)(1 -tan(x)²). It was multiplying by tan(x) that made the π/2, 3π/2 solutions disappear. Had we kept 1/tan(x) as a numerator factor of cot(x), we would have had the same result as seen in the second approach.