Solving Equations with Inverse Cosine

  • 11th Grade
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Cierra Henderson, MBA |
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Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
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Quizzes Created: 8156 | Total Attempts: 9,588,805
| Questions: 20 | Updated: Jan 19, 2026
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1) Solve for θ in [0, 2π]: cos(θ) = −√2/2

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About This Quiz
Solving Equations With Inverse Cosine - Quiz

Now it’s time to put inverse cosine to work in equation solving. Here, you’ll solve trigonometric equations step by step, finding all solutions for θθθ within the interval [0,2π][0, 2π][0,2π]. This quiz will test your ability to handle reflections, phase shifts, and multiple solutions while keeping everything locked within the... see moreproper domain.
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2) Solve for θ in [0, 2π]: cos(θ) = 1/2

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3) Solve for θ in [0, 2π]: cos(θ) = √3/2

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4) Solve for θ in [0, 2π]: cos(θ) = −1/2

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5) Solve for θ in [0, 2π]: cos(θ) = 0

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6) Solve for θ in [0, 2π]: cos(θ) = −√3/2

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7) Solve for θ in [0, 2π]: cos(2θ) = 1/2

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8) Solve for θ in [0, 2π]: cos(θ − π/3) = 0

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9) Solve for θ in [0, 2π]: cos(θ + π/6) = −1/2

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10) Solve for θ in [0, 2π]: cos(2θ) = −√2/2

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11) Solve for θ in [0, 2π]: cos(θ) = −1

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12) Solve for θ in [0, 2π]: cos(3θ) = 1

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13) Solve for θ in [0, 2π]: cos(θ) = (√6 + √2)/4

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14) Solve for θ in [0, 2π]: cos(θ) = (√6 − √2)/4

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15) Solve for θ in [0, 2π]: cos(θ) = −(√6 − √2)/4

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16) Solve for θ in [0, 2π]: cos(θ) = −(√6 + √2)/4

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17) Solve for θ in [0, 2π]: cos(θ + π/4) = √2/2

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18) Solve for θ in [0, 2π]: cos(θ − π/2) = 0

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19) Solve for θ in [0, 2π]: cos(2θ − π/3) = −1

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20) Solve for θ in [0, 2π]: cos(θ) = √2/2

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Cierra Henderson |MBA |
K-12 Expert
Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
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Solve for θ in [0, 2π]: cos(θ) = −√2/2
Solve for θ in [0, 2π]: cos(θ) = 1/2
Solve for θ in [0, 2π]: cos(θ) = √3/2
Solve for θ in [0, 2π]: cos(θ) = −1/2
Solve for θ in [0, 2π]: cos(θ) = 0
Solve for θ in [0, 2π]: cos(θ) = −√3/2
Solve for θ in [0, 2π]: cos(2θ) = 1/2
Solve for θ in [0, 2π]: cos(θ − π/3) = 0
Solve for θ in [0, 2π]: cos(θ + π/6) = −1/2
Solve for θ in [0, 2π]: cos(2θ) = −√2/2
Solve for θ in [0, 2π]: cos(θ) = −1
Solve for θ in [0, 2π]: cos(3θ) = 1
Solve for θ in [0, 2π]: cos(θ) = (√6 +...
Solve for θ in [0, 2π]: cos(θ) = (√6 −...
Solve for θ in [0, 2π]: cos(θ) = −(√6 − √2)/4
Solve for θ in [0, 2π]: cos(θ) = −(√6 + √2)/4
Solve for θ in [0, 2π]: cos(θ + π/4) = √2/2
Solve for θ in [0, 2π]: cos(θ − π/2) = 0
Solve for θ in [0, 2π]: cos(2θ − π/3) = −1
Solve for θ in [0, 2π]: cos(θ) = √2/2
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