Learn how to solve a multi step equation using trig

Learn how to solve a multi step equation using trig

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

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Quizizz Content

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The video tutorial covers solving a mathematical equation involving secant and cosine functions. It begins with an introduction to the problem and common student mistakes. The instructor demonstrates solving the equation, emphasizing the importance of considering plus or minus signs. Rationalizing the denominator and finding cosine values are discussed, followed by simplifying solutions using π and 2π. The tutorial concludes with a focus on ensuring all possible solutions are named and addresses student questions.

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7 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in solving the equation presented in the video?

Subtract 4 from both sides

Add 4 to both sides

Multiply both sides by 3

Divide both sides by 3

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is it important to consider both positive and negative solutions when taking the square root?

To ensure all possible solutions are found

To simplify the equation

To make the equation easier to solve

To eliminate complex numbers

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the reciprocal of the secant function?

Sine

Cosine

Tangent

Cosecant

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which angles on the unit circle correspond to a cosine value of plus or minus radical 3/2?

π/4, 3π/4, 5π/4, 7π/4

π/6, 5π/6, 7π/6, 11π/6

π/2, 3π/2, π, 2π

π/3, 2π/3, 4π/3, 5π/3

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the benefit of adding π instead of 2π when listing all possible solutions?

It increases the number of solutions

It eliminates negative solutions

It simplifies the equation

It reduces redundancy

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How many times can π be added to find all solutions?

Once

Twice

Infinite times

Three times

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main goal when naming all possible solutions?

To ensure all solutions are accounted for

To reduce the number of solutions

To eliminate complex numbers

To find the simplest solution