Solving 1st and 2nd Degree Trigonometric Equations

Solving 1st and 2nd Degree Trigonometric Equations

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Flashcard

Mathematics

11th Grade

Hard

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15 questions

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1.

FLASHCARD QUESTION

Front

What is the general form of a 1st degree trigonometric equation?

Back

A 1st degree trigonometric equation is an equation that can be expressed in the form of a trigonometric function equal to a constant, such as sin(θ) = k, cos(θ) = k, or tan(θ) = k.

2.

FLASHCARD QUESTION

Front

What is the general form of a 2nd degree trigonometric equation?

Back

A 2nd degree trigonometric equation is an equation that can be expressed in the form of a trigonometric function squared equal to a constant or another trigonometric function, such as sin²(θ) = k or 1 + tan²(θ) = sec²(θ).

3.

FLASHCARD QUESTION

Front

What is the range of the secant function?

Back

The range of the secant function, sec(θ), is (-∞, -1] ∪ [1, ∞).

4.

FLASHCARD QUESTION

Front

What is the range of the cosecant function?

Back

The range of the cosecant function, csc(θ), is (-∞, -1] ∪ [1, ∞).

5.

FLASHCARD QUESTION

Front

How do you solve the equation -1 - 2sec²(θ) = -3sec²(θ)?

Back

Rearranging gives sec²(θ) = 1, leading to θ = 0, π within the interval [0, 2π].

6.

FLASHCARD QUESTION

Front

What is the solution to the equation -2 = -5 + 3sec(θ)?

Back

Rearranging gives sec(θ) = 1, leading to θ = 0 within the interval [0, 2π].

7.

FLASHCARD QUESTION

Front

Is it true that sin(θ) = -1 has no solutions?

Back

FALSE; sin(θ) = -1 has solutions, specifically θ = 3π/2.

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