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Algebra, MET2 2019 VCAA 19 MC

Given that  `tan(alpha) = d`, where  `d > 0`  and  `0 < alpha < pi/2`, the sum of the solutions to  `tan (2x) = d`, where  `0 < x < (5 pi)/4`, in terms of  `alpha`  is

  1. `0`
  2. `2 alpha`
  3. `pi + 2 alpha`
  4. `pi/2 + alpha`
  5. `(3 (pi + alpha))/2`
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`E`

Show Worked Solution

`tan(alpha) = tan(2x) = d`

`text(Period of)\ \ tan alpha = pi`

`2x = alpha, quad alpha + pi, quad alpha + 2 pi`

`x = alpha/2, quad (alpha + pi)/2, quad (alpha + 2 pi)/2`

`text(Given)\ \ 0 < alpha < pi/2, => (alpha + 2 pi)/2 < (5 pi)/4`

`:. sum text(solutions)` `= alpha/2 + (alpha + pi)/2 + (alpha + 2 pi)/2`
  `= (3 (pi + alpha))/2`

 
`=>   E`

Algebra, MET1 2009 VCAA 4

Solve the equation  `tan (2x) = sqrt 3`  for  `x in (– pi/4, pi/4) uu (pi/4, (3 pi)/4).`  (3 marks)

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`:. x = pi/6, (2 pi)/3`

Show Worked Solution

`tan (2x) = sqrt 3`

`=>\ text(Base angle)\ = pi/3`

`2x` `= pi/3, (4pi)/3, (-2pi)/3, …`
`:.x` `= pi/6 + (2pi)/3, (-pi)/3, …`
  `=pi/6 or (2 pi)/3,\ \ \ \ x in (– pi/4, pi/4) uu (pi/4, (3 pi)/4)`