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Functions, MET1 2023 VCE SM-Bank 3

Find the general solution for  \(2 \sin (x)=\tan (x)\) for \(x \in R\).  (3 marks)

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\(x=n\pi\ ,\ \Big(2n\pm \dfrac{\pi}{3}\Big)\pi\quad n\in \mathbb{Z}\)

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\(2\sin(x)\) \(=\tan(x)\)
\(2\sin(x)\) \(=\dfrac{\sin(x)}{\cos(x)}\quad \cos(x)\neq 0\)
\(2\sin(x)\cos(x)-\sin(x)\) \(=0\)
\(\sin(x)\Big(2\cos(x)-1\Big)\) \(=0\)
\(\therefore \sin(x)=0\quad\) \(\text{or}\) \(\quad 2\cos(x)-1=0\)
\(\therefore x=n\pi\quad\) \(\text{or}\) \(\quad \cos(x)=\dfrac{1}{2}\)
    \(\quad x=2n\pi\pm \dfrac{\pi}{3}=\Bigg(2n\pm \dfrac{1}{3}\Bigg)\pi\quad n\in\mathbb{Z}\)

Algebra, MET2 2020 VCAA 4 MC

The solutions of the equation  `2cos(2x-(pi)/(3))+1=0`  are

  1. `x=(pi(6k-2))/(6)" or "x=(pi(6k-3))/(6)," for "k in Z`
  2. `x=(pi(6k-2))/(6)" or "x=(pi(6k+5))/(6)," for "k in Z`
  3. `x=(pi(6k-1))/(6)" or "x=(pi(6k+2))/(6)," for "k in Z`
  4. `x=(pi(6k-1))/(6)" or "x=(pi(6k+3))/(6)," for "k in Z`
  5. `x=pi" or "x=(pi(6k+2))/(6)," for "k in Z`
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`D`

Show Worked Solution
`2cos(2x-(pi)/(3))+1` `=0`  
`cos(2x-(pi)/(3))` `=- 1/2`  
`2x-(pi)/(3)` `=(2pi)/3\ \ text(or)\ \ -(2pi)/3`  

 
`text(General Solution:)`

`2x-(pi)/(3)` `=2kpi+(2pi)/3`  
`2x` `=2kpi+pi`  
`x` `=kpi+pi/2`  
  `=pi/6(6k+3)`  

 

`2x-(pi)/(3)` `=2kpi-(2pi)/3`  
`2x` `=2kpi-pi/3`  
`x` `=kpi-pi/6`  
  `=pi/6(6k-1)`  

`=>D`

Functions, MET1 2021 VCAA 3

Consider the function  `g: R -> R, \ g(x) = 2sin(2x).`

  1. State the range of `g`.  (1 mark)
  2. State the period of `g`.  (1 mark)
  3. Solve  `2 sin(2x) = sqrt3`  for  `x ∈ R`.  (3 marks)
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  1. `[–2,2]`
  2. `pi`
  3. `x= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`
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a.   `text(S)text(ince)  -1<sin(2x)<1,`

`text(Range)\  g(x) = [–2,2]`
 

b.   `text(Period) = (2pi)/n = (2pi)/2 = pi`
 

c.    `2sin(2x)` `=sqrt3`
  `sin(2x)` `=sqrt3/2`
  `2x` `=pi/3, (2pi)/3, pi/3 + 2pi, (2pi)/3 + 2pi, …`
  `x` `=pi/6, pi/3, pi/6+pi, pi/3+pi, …`

 
`:.\ text(General solution)`

`= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`

Algebra, MET2 2009 VCAA 4 MC

The general solution to the equation  `sin (2x) = -1`  is

  1. `x = n pi - pi/4,\ n in Z`
  2. `x = 2n pi + pi/4 or x = 2n pi - pi/4,\ n in Z`
  3. `x = (n pi)/2 + (-1)^n pi/2,\ n in Z`
  4. `x = (n pi)/2 + (-1)^n pi/4,\ n in Z`
  5. `x = n pi + pi/4 or x = 2n pi + pi/4,\ n in Z`
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`A`

Show Worked Solution
`2x` `= 2n pi – pi/2,\ \ n in Z`
`x` `= n pi – pi/4,\ \ n in Z`

 
`=>   A`