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Trigonometry, 2ADV T2 2023 HSC 20

Find all the values of `theta`, where  `0^@ <=theta <= 360^@`, such that

`sin(theta-60^@)=-sqrt3/2`  (3 marks)

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`theta=0^@, 300^@ and 360^@`

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`sin60^@=sqrt3/2\ \ =>\ \ text{Base angle}\ =60^@`

`=>\ text{sin is negative in 3rd and 4th quadrants}`

`sin(theta-60^@)` `=180+60, 360-60`  
  `=240^@, 300^@`  

 
`theta-60^@=240^@\ \ =>\ \ theta=300^@`

`theta-60^@=300^@\ \ =>\ \ theta=360^@`

 
`text{Consider}\ theta = 0^@`

`sin(0-60^@)=sin(-60^@)=-sqrt3/2`

`:.theta=0^@, 300^@ and 360^@`

Trigonometry, 2ADV T2 SM-Bank 39

Let  `f(x) = sin((2pix)/3)`.

Solve the equation  `sin((2pix)/3) = -sqrt3/2`  for  `0<=x<=3`  (2 marks)

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`x = 2, 5/2`

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`sin((2pix)/3) = -sqrt3/2`

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

`(2 pi x)/3` `=(4pi)/3, (5pi)/3, (10pi)/3, …` 
`:.x` `=2 or 5/2, \ \ \ (0<=x<=3)`

Trigonometry, 2ADV T2 SM-Bank 36

Solve the equation  `sin (x/2) = -1/2`  for  `2 pi<=x<= 4 pi`  (2 marks)

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`x = (7 pi)/3, (11 pi)/3`

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`x/2` `=pi/6 + pi, 2pi – pi/6, 2pi + (pi/6 +pi), …`
  `=(7pi)/6, (11pi)/6, (19pi)/6, …`
`:. x` `=(7pi)/3, (11pi)/3, (19pi)/3, …`
   

`text(Given)\ \ \2 pi<=x<= 4 pi`

`:. x = (7 pi)/3, (11 pi)/3`

Trigonometry, 2ADV T2 SM-Bank 35

Solve the equation

`sin (2x + pi/3) = 1/2\ \ text(for)\ \ 0<= x <=pi`  (2 marks)

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`x = pi/4, (11 pi)/12`

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`sin (2x + pi/3) = 1/2`

`=>\ text(Base angle is)\ \ pi/6`

`(2x + pi/3)` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, …`
`2x` `= – pi/6, pi/2, (11pi)/6, (15pi)/6, …`
`x` `= – pi/12, pi/4, (11pi)/12, (15pi)/12, …`

 
`:. x = pi/4, (11 pi)/12\ \ (0<=x<=pi)`

Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of  `2 sin theta = 1`  for  `0 <= theta <= 2 pi`.  (2 marks)

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`pi/6, (5 pi)/6`

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`2 sin theta` `= 1,\ \ \ \ 0 <= theta <= 2 pi`
`sin theta` `= 1/2`

`=> sin (pi/6) = 1/2 \ \ text(and sin is positive)`

`text(in the 1st/2nd quadrants)`

`:. theta` `= pi/6, pi – pi/6`
  `= pi/6, (5 pi)/6`

Trigonometry, 2ADV T2 2008 HSC 6a

Solve  `2 sin^2 (x/3) = 1`  for  `-pi <= x <= pi`.   (3 marks)

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`x = -(3pi)/4, (3pi)/4`

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♦♦ Although exact data not available, markers specifically mentioned this question was poorly answered.
`2 sin^2 (x/3)` `= 1\ \ text(for)\ \ -pi <= x <= pi`
`sin^2 (x/3)` `= 1/2`
`sin (x/3)` `= +- 1/sqrt2`

 

MARKER’S COMMENT: Many students had problems adjusting their answer to the given domain, especially when dealing with negative angles.

`text(When)\ sin (x/3) = 1/sqrt2`

`x/3` `= pi/4, (3pi)/4`
`x` `= (3pi)/4, (9pi)/4`

 

`text(When)\ sin (x/3) = – 1/sqrt2`

`x/3` `= – pi/4, -(3pi)/4`
`x` `= -(3pi)/4,  -(9pi)/4`

 

`:.\ x = -(3pi)/4\ \ text(or)\ \  (3pi)/4\ \ text(for)\ \ -pi <= x <= pi` 

Trigonometry, 2ADV T2 2011 HSC 2b

Find the exact values of  `x`  such that  `2sin x = - sqrt3`, where  `0 <= x <= 2pi`.   (2 marks)

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 `x = (4pi)/3,\ (5pi)/3`

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MARKER’S COMMENT: Better responses found the reference angle and then identified the correct quadrants, as shown clearly in the worked solution.
`2sinx` `=- sqrt3\ \ text(where)\ \ 0 <= x <= 2pi`
`sin x` `= -sqrt3/2`
`sin (pi/3)` `= sqrt3/2`

 

`text(S)text(ince)\ sin x\ text(is negative in)\ 3^text(rd) // 4^text(th)\ text(quadrants)`

`x` `= pi + pi/3,\ 2pi\ – pi/3`
  `= (4pi)/3,\ (5pi)/3\ \ text(radians)`