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Find all the values of θ, where 0° ≤ θ ≤ 360°, such that $$ ext{sin}( heta - 60°) = rac{ ext{√}3}{2} - HSC - SSCE Mathematics Advanced - Question 20 - 2023 - Paper 1

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Find-all-the-values-of-θ,-where-0°-≤-θ-≤-360°,-such-that--$$--ext{sin}(-heta---60°)-=--rac{-ext{√}3}{2}-HSC-SSCE Mathematics Advanced-Question 20-2023-Paper 1.png

Find all the values of θ, where 0° ≤ θ ≤ 360°, such that $$ ext{sin}( heta - 60°) = rac{ ext{√}3}{2}. $$

Worked Solution & Example Answer:Find all the values of θ, where 0° ≤ θ ≤ 360°, such that $$ ext{sin}( heta - 60°) = rac{ ext{√}3}{2} - HSC - SSCE Mathematics Advanced - Question 20 - 2023 - Paper 1

Step 1

Recognises that sin60° = √3 / 2

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Answer

We know from trigonometric identities that:

ext{sin}(60°) = rac{ ext{√}3}{2}.

This provides us with a basis for solving the equation.

Step 2

−60° ≤ θ − 60° ≤ 300°

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Answer

We will set up the equations as follows:

heta−60°=60°+nimes360°extorheta−60°=120°+nimes360°. heta - 60° = 60° + n imes 360° ext{ or } heta - 60° = 120° + n imes 360°.

Solving for θ gives:

  1. Case 1:
heta−60°=60°ightarrowheta=120° heta - 60° = 60° ightarrow heta = 120°
  1. Case 2:
heta−60°=120°ightarrowheta=180° heta - 60° = 120° ightarrow heta = 180°
  1. Case 3:
heta−60°=240°ightarrowheta=300° heta - 60° = 240° ightarrow heta = 300°
  1. Case 4:
heta−60°=300°ightarrowheta=360°. heta - 60° = 300° ightarrow heta = 360°.

Step 3

θ = 300°, 360°, 120°

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Answer

Thus, the final solutions for θ, considering the range of 0° ≤ θ ≤ 360°, are:

θ=120°,300°,360°.θ = 120°, 300°, 360°.

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