Even Odd Equations Quiz: Solving Equations Using Even-Odd Properties

  • 11th Grade
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Quizzes Created: 7682 | Total Attempts: 9,547,133
| Questions: 20 | Updated: Dec 16, 2025
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1) Solve on [0, 2π]: sin(−θ) = −√2/2

Explanation

sin is odd: sin(−θ)=−sinθ. Then −sinθ=−√2/2 ⇒ sinθ=√2/2. Solutions to sinθ=√2/2 are θ=π/4, 3π/4. The values satisfying the original equation are θ whose negatives equal those angles: θ≡−π/4, −3π/4 (mod 2π) ⇒ θ=7π/4, 5π/4.

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About This Quiz
Even Odd Equations Quiz: Solving Equations Using Even-odd Properties - Quiz

Ready to apply even–odd properties to solve trigonometric equations more effectively? In this quiz, you’ll use symmetry-based reasoning to rewrite expressions, analyze sign behavior, and uncover solutions that might otherwise be hidden. You’ll practice identifying when functions flip or remain unchanged, helping you simplify equations before solving them. Through step-by-step... see moreproblems, you’ll develop confidence using even–odd logic as a strategic shortcut that strengthens both accuracy and efficiency in trigonometric problem solving.
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2) Select all θ in [0, 2π] satisfying cos(−θ) = cosθ

Explanation

cos is even for all inputs, so cos(−θ)=cosθ holds for every θ where cosine is defined.

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3) Solve on [0, 2π]: tan(−θ) = √3

Explanation

tan is odd: tan(−θ) = −tanθ. So −tanθ = √3 ⇒ tanθ = −√3. On [0,2π], tanθ = −√3 at θ = 5π/6 and 11π/6.

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4) For all θ where defined, tan(−θ) = −tanθ.

Explanation

tanθ = sinθ/cosθ with sin odd and cos even, so tan(−θ) = sin(−θ)/cos(−θ) = (−sinθ)/(cosθ) = −tanθ.

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5) Solve on [0, 2π]: cos(−θ) = −1

Explanation

cos is even: cos(−θ) = cosθ. Solve cosθ = −1, whose solution on [0, 2π] is θ = π.

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6) On [−π, π], if sin(−θ) = b, then θ solves sinθ = −b.

Explanation

sin is odd: sin(−θ)=−sinθ. So sin(−θ)=b ⇔ −sinθ=b ⇔ sinθ=−b, then find θ in [−π, π] accordingly.

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7) Select all solutions on [0, 2π]: tan(−θ) = −1

Explanation

tan odd ⇒ −tanθ = −1 ⇒ tanθ = 1. On [0,2π], tanθ = 1 at θ = π/4 and 5π/4. Thus tan(−θ) = −1 holds for θ = π/4, 5π/4.

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8) Solve on [0, 2π]: cos(−θ) = √3/2

Explanation

cos is even: cos(−θ)=cosθ. Solve cosθ = √3/2. On [0, 2π], solutions are θ = π/6 and 11π/6.

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9) Select all solutions on [0, 2π]: sin(−θ) = 0

Explanation

sin is odd: sin(−θ) = −sinθ. Then −sinθ = 0 ⇒ sinθ = 0. On [0, 2π], sinθ = 0 at θ = 0, π, 2π.

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10) Solve on [−π, π]: tan(−θ) = 0

Explanation

tan is odd: tan(−θ)=−tanθ. Then −tanθ=0 ⇒ tanθ=0. On [−π, π], the unique solution that equals its own negative is θ=0.

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11) Select all solutions on [−π, π]: cos(−θ) = 0

Explanation

cos is even: cos(−θ)=cosθ. Solve cosθ=0. On [−π, π], cosθ=0 at θ=−π/2 and θ=π/2.

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12) Solve on [−π, π]: sin(−θ) = −1/2

Explanation

Oddness gives −sinθ = −1/2 ⇒ sinθ = 1/2. On [−π, π], sinθ = 1/2 at θ = π/6 and 5π/6. But we need θ such that sin(−θ)=−1/2, which came from sinθ=1/2, so θ = −π/6 and −5π/6 satisfy the original equation.

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13) On [0, 2π], if cos(−θ) = a, then θ solves cosθ = a.

Explanation

cos is even, so cos(−θ)=cosθ. Solving cos(−θ)=a is equivalent to solving cosθ=a on the same interval.

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14) Solve on [−π, π]: cos(−θ) = −1/2

Explanation

cos even ⇒ cosθ = −1/2. On [−π, π], cosθ = −1/2 at θ = ±2π/3.

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15) Solve on [0, 2π]: sin(−θ) = −1

Explanation

sin is odd: sin(−θ) = −sinθ. Equation becomes −sinθ = −1 ⇒ sinθ = 1. On [0, 2π], θ = π/2.

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16) Solve on [0, 2π]: sin(−θ) = 1/2

Explanation

Use oddness: sin(−θ) = −sinθ. Then −sinθ = 1/2 ⇒ sinθ = −1/2. On [0, 2π], sinθ = −1/2 at θ = 7π/6 and 11π/6.

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17) Solve on [−π, π]: sin(−θ) = 0 and θ ≥ 0

Explanation

sin odd ⇒ −sinθ = 0 ⇒ sinθ = 0. On [−π, π] with θ ≥ 0, the solutions are θ = 0 and θ = π.

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18) Solve on [0, 2π]: sin(−θ) = sinθ

Explanation

sin is odd, so sin(−θ) = −sinθ. Equality sin(−θ)=sinθ implies −sinθ=sinθ ⇒ 2sinθ=0 ⇒ sinθ=0. On [0,2π], θ=0, π, 2π.

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19) If sin(−θ) = a has two solutions in [0, 2π], then sinθ = −a will have the same two solutions in [0, 2π].

Explanation

sin(−θ)=a ⇔ −sinθ=a ⇔ sinθ=−a. The solution sets coincide within the same interval boundaries.

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20) For all θ, cos(−θ) = −cosθ.

Explanation

cos is even, so cos(−θ) = cosθ, not −cosθ.

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Solve on [0, 2π]: sin(−θ) = −√2/2
Select all θ in [0, 2π] satisfying cos(−θ) = cosθ
Solve on [0, 2π]: tan(−θ) = √3
For all θ where defined, tan(−θ) = −tanθ.
Solve on [0, 2π]: cos(−θ) = −1
On [−π, π], if sin(−θ) = b, then θ solves sinθ = −b.
Select all solutions on [0, 2π]: tan(−θ) = −1
Solve on [0, 2π]: cos(−θ) = √3/2
Select all solutions on [0, 2π]: sin(−θ) = 0
Solve on [−π, π]: tan(−θ) = 0
Select all solutions on [−π, π]: cos(−θ) = 0
Solve on [−π, π]: sin(−θ) = −1/2
On [0, 2π], if cos(−θ) = a, then θ solves cosθ = a.
Solve on [−π, π]: cos(−θ) = −1/2
Solve on [0, 2π]: sin(−θ) = −1
Solve on [0, 2π]: sin(−θ) = 1/2
Solve on [−π, π]: sin(−θ) = 0 and θ ≥ 0
Solve on [0, 2π]: sin(−θ) = sinθ
If sin(−θ) = a has two solutions in [0, 2π], then sinθ = −a...
For all θ, cos(−θ) = −cosθ.
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