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Lesson: Solving a Trigonometric Equation by Squaring

Worksheet • 7 Questions

Q1:

Consider the equation s i n c o s πœƒ + πœƒ = √ 2 , where 0 < πœƒ ≀ 3 6 0 ∘ ∘ . Call this Equation A.

Create Equation B by squaring both sides of Equation A. Use the fact that s i n c o s 2 2 πœƒ + πœƒ = 1 to simplify Equation B.

  • A 2 πœƒ πœƒ = 1 s i n c o s
  • B s i n c o s πœƒ πœƒ = βˆ’ 1
  • C s i n c o s πœƒ πœƒ = 1
  • D 2 πœƒ πœƒ = βˆ’ 1 s i n c o s
  • E s i n c o s πœƒ πœƒ = 2

Now, use a double angle formula to further simplify Equation B.

  • A s i n 2 πœƒ = 1
  • B s i n 2 πœƒ = 2
  • C s i n πœƒ = 1
  • D c o s πœƒ = 1
  • E c o s 2 πœƒ = 1

The solutions to Equation A are a subset of the solutions of Equation B. Using this, solve Equation A over the specified range.

  • A πœƒ = 4 5 ∘
  • B πœƒ = 2 1 5 ∘
  • C πœƒ = 3 0 ∘
  • D πœƒ = 1 3 5 ∘
  • E πœƒ = 6 0 ∘

Q2:

Solve √ 2 πœƒ + √ 3 πœƒ = 2 s i n c o s , where 0 < πœƒ ≀ 2 πœ‹ . Give your answer in radians to three significant figures.

  • A πœƒ = 0 . 2 2 1 , 1 . 1 5
  • B πœƒ = 0 . 3 9 6 , 2 . 9 5
  • C πœƒ = 0 . 4 7 1 , 2 . 1 7
  • D πœƒ = 0 . 2 4 1 , 1 . 8 6
  • E πœƒ = 1 . 6 9 , 2 . 1 4

Q3:

If πœƒ ∈ [ 0 , 1 8 0 [ ∘ ∘ and s i n c o s πœƒ + πœƒ = 1 , find the possible values of πœƒ .

  • A 0 ∘ , 9 0 ∘
  • B 0 ∘ , 4 5 ∘
  • C 0 ∘ , 1 8 0 ∘
  • D 4 5 ∘ , 9 0 ∘
  • E 9 0 ∘ , 1 8 0 ∘

Q4:

By first squaring both sides, or otherwise, solve the equation 4 πœƒ βˆ’ 4 πœƒ = √ 3 s i n c o s , where 0 < πœƒ ≀ 3 6 0 . Be careful to remove any extraneous solutions. Give your answers to two decimal places.

  • A πœƒ = 6 2 . 8 3 , 2 0 7 . 1 7 ∘ ∘
  • B πœƒ = 7 7 . 2 4 , 2 1 0 . 5 7 ∘ ∘
  • C πœƒ = 8 6 . 1 4 , 2 1 2 . 5 7 ∘ ∘
  • D πœƒ = 4 7 . 3 5 , 1 9 5 . 1 2 ∘ ∘
  • E πœƒ = 6 5 . 1 8 , 2 0 5 . 1 4 ∘ ∘

Q5:

Find the set of values satisfying 9 7 πœƒ + 6 0 πœƒ = 0 s i n c o s where 0 < πœƒ < 3 6 0 ∘ ∘ . Give the answers to the nearest second.

  • A { 1 4 8 1 5 β€² 3 9 β€² β€² , 3 2 8 1 5 β€² 3 9 β€² β€² } ∘ ∘
  • B { 3 1 4 4 β€² 2 1 β€² β€² , 1 4 8 1 5 β€² 3 9 β€² β€² } ∘ ∘
  • C { 3 1 4 4 β€² 2 1 β€² β€² , 3 2 8 1 5 β€² 3 9 β€² β€² } ∘ ∘
  • D { 3 1 4 4 β€² 2 1 β€² β€² , 2 1 1 4 4 β€² 2 1 β€² β€² } ∘ ∘
  • E { 1 4 8 1 5 β€² 3 9 β€² β€² , 2 1 1 4 4 β€² 2 1 β€² β€² } ∘ ∘

Q6:

Find the set of values satisfying s i n c o s πœƒ βˆ’ πœƒ = 0 where 0 < πœƒ < 3 6 0 ∘ ∘ . Give the answers to the nearest second.

  • A { 4 5 0 β€² 0 β€² β€² , 2 2 5 0 β€² 0 β€² β€² } ∘ ∘
  • B { 2 2 5 0 β€² 0 β€² β€² , 1 3 5 0 β€² 0 β€² β€² } ∘ ∘
  • C { 4 5 0 β€² 0 β€² β€² , 1 3 5 0 β€² 0 β€² β€² } ∘ ∘
  • D { 2 2 5 0 β€² 0 β€² β€² , 3 1 5 0 β€² 0 β€² β€² } ∘ ∘
  • E { 4 5 0 β€² 0 β€² β€² , 3 1 5 0 β€² 0 β€² β€² } ∘ ∘

Q7:

If πœƒ ∈ ( 1 8 0 , 3 6 0 ) ∘ ∘ and s i n c o s πœƒ + πœƒ = βˆ’ 1 , find the value of πœƒ .

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