How to apply inverse operations to find all of the solutions to an equation with cosine

How to apply inverse operations to find all of the solutions to an equation with cosine

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Wayground Content

FREE Resource

The video tutorial covers solving trigonometric equations, focusing on cosine squared equations. It explains using variables to simplify the process and highlights the importance of the unit circle in finding solutions. The tutorial discusses finding solutions in different quadrants and introduces the concept of periodicity to determine general solutions. It concludes with methods to simplify solutions using periodic properties.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in solving the equation 4X^2 - 3 = 0?

Subtract 3 from both sides

Add 3 to both sides

Divide both sides by 4

Multiply both sides by 4

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

After solving 4X^2 = 3, what is the value of X?

± sqrt(2)/4

± sqrt(3)/2

± sqrt(2)/3

± sqrt(3)/4

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which angle in the first quadrant has a cosine value of sqrt(3)/2?

π/3

π/2

π/4

π/6

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the significance of the unit circle in trigonometry?

It helps in solving algebraic equations

It represents angles as coordinates

It is used to measure angles in degrees

It is a tool for graphing linear equations

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How many solutions are there for the equation between 0 and 2π?

Three

Four

Two

Five

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is a coterminal angle?

An angle that shares the same terminal side

An angle that is less than 90 degrees

An angle that is a multiple of π

An angle that is greater than 180 degrees

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How can the expression of solutions be simplified using periodicity?

By adding π to each solution

By subtracting π from each solution

By adding 2π to each solution

By multiplying each solution by π

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