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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 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 2016 HSC 11g

Solve  `sin (x/2) = 1/2`  for  `0 <= x <= 2pi.`  (2 marks)

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`x = pi/3,\ \ (5 pi)/3`

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

`=>\ text(Reference angle) = pi/6,`

`x/2` `= pi/6, pi-pi/6, 2pi + pi/6, …`
  `= pi/6, (5 pi)/6, (13 pi)/6, …`

 

`:. x = pi/3,\ \ (5 pi)/3\ \ text(for)\ \ 0 <= x <= 2 pi`

Trigonometry, 2ADV T2 2007 HSC 4a

Solve  `sqrt 2\ sin\ x = 1`  for  `0 <= x <= 2 pi`.  (2 marks)

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`pi/4 ,  (3 pi)/4`

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`sqrt 2\ sin\ x` `= 1` `\ \ \ \ \ \ \ 0 <= x <= 2 pi`
`sin\ x` `= 1/sqrt 2`  

`=>sin\ pi/4 = 1/sqrt 2\ \ text(and sin is positive in)`

`text(1st/2nd quadrants,)`

`:. x` `= pi/4\ ,\ pi – pi/4`
  `= pi/4\ ,\ (3 pi)/4`

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)`