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Worksheets

Steps to Solve Trig Equations

Total questions: 14

Worksheet time: 13mins

Name
Class
Date
1.

Which of these is equivalent to 2cos2x − 3cosx = 0 ?

a)

-cos2x = 0

b)

cosx(2cosx + 3) = 0

c)

cosx(2cosx − 3) = 0

d)

cos x = ⅔

2.

Which is a correct way to solve the equation

cosx(2cosx − 3) = 0?

a)

divide cos x from both sides

b)

set each factor equal to 0 and solve

c)

distribute cosx into the parentheses

d)

guess and hope for the best

3.

Why doesn't 2cosx − 3 = 0 have solutions?

a)

cos x is never bigger than one

b)

cos x is never equal to a fraction

c)

Actually, this equation does have a solution, x = π

d)

This equation will have a solution tomorrow.

4.

Choose a good way to start solving this equation

tanx sin2x = 2 tanx

a)

divide tanx from both sides

b)

factor out tanx

c)

subtract 2 tanx from the left and then factor tanx out

d)

cancel sin2x out

5.

So tanx sin2x = 2 tanx is equivalent to tanx(sin2x − 2) = 0

Then tanx = 0 or sin2x − 2 = 0, now what?

a)

solutions come only from tanx = 0,

b)

solutions come only from sin2x − 2 = 0

c)

solutions come from where sin x is undefined

d)

solutions come from where tan x is undefined

6.

The square root of 2 is bigger than 1.

a)

true

b)

false

7.

To solve this equation, cos2x + sinx = 1, replace cos2x with

a)

1/(sec2x)

b)

sin2x − 1

c)

1 − sin2x

d)

1 + tan2x

8.

Organize this equation so it can be factored:

1−sin2x + sinx = 1

a)

move all to the right: 0 = sin2x − sinx

b)

cancel the ones and cancel a sinx

c)

replace sin2x with 1 − cos2x

d)

use this: 0 = sin2x − sinx + 2

9.

Factor 0 = sin2x − sinx

a)

0 = sinx(sinx)

b)

0 = sinx(1 − sinx)

c)

0 = cosx(sinx − 1)

d)

0 = sinx(sinx − 1)

10.

On the domain [0, 2π), solve this equation 0 = sinx(sinx − 1)

a)

0, π

b)

0, π, ½π

c)

½π

d)

0, 90, 270

11.

csc2x = 2 is equivalent to sin2x = ½

a)

True

b)

False

12.

To solve the equation sec2x − secx = 2

a)

start by subtracting 2 from both sides and then factor

b)

add sec x to both sides and factor

c)

factor out secx and set each factor equal to 2

d)

sec x is never bigger than 1, so there are no solutions

13.

Factor: sec2x − secx − 2

a)

(sec x)(secx − 2)

b)

(secx − 2)(secx − 1)

c)

(secx − 2)(secx + 1)

d)

(secx + 2)(secx − 1)

14.

Solve the equation (secx − 2)(secx + 1) = 0 on [0, 3600)

a)

60⁰, 180⁰, 300⁰

b)

30⁰, 180⁰,330⁰

c)

60⁰, 120⁰

d)

180⁰