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Quiz Solutions for MTH 120 - Summer 2005 - Essex County College, Quizzes of Pre-Calculus

The solutions to quiz # 81 for the mth 120 course at essex county college, held during the summer 2005 semester. The quiz covers problems related to solving trigonometric equations in the interval [0, 2π).

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Pre 2010

Uploaded on 08/08/2009

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MTH 120 Summer 2005
Essex County College Division of Mathematics
Quiz # 81 Printed June 9, 2005
Name:
Signature:
Show all work clearly and in order, and box your final answers . Justify your answers alge-
braically whenever possible. You have 20 minutes to take this 10 point quiz. When you do use
your calculator, sketch all relevant graphs and write down all relevant mathematics.
1. Solve for x.
(a) sin x+2 = sin xin the interval [0,2π).
Solution:
sin x+2 = sin x
2 sin x=2
sin x=
2
2.
The reference is 45and the solutions occur in the third and fourth quadrant, so
x= 45+ 180= 225=5π
4and x= 360
45= 315=7π
4
(b) 2 sin2xsin x1 = 0 in the interval [0,2π).
Solution:
2 sin2xsin x1=0
(2 sin x+ 1) (sin x1) = 0.
Which gives two equations to solve.
sin x=
1
2and sin x= 1.
The reference for the first equation is 30and the solutions occur in the third and
fourth quadrant, so
x= 30+ 180= 210=7π
6and x= 360
30= 330=11π
6,
and the solution to the second equation is x=π
2
1This document was prepared by Ron Bannon using L
A
T
E
X 2ε.
1

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MTH 120 — Summer — 2005 Essex County College — Division of Mathematics

Quiz # 8^1 — Printed June 9, 2005

Name:

Signature:

Show all work clearly and in order, and box your final answers. Justify your answers alge-

braically whenever possible. You have 20 minutes to take this 10 point quiz. When you do use

your calculator, sketch all relevant graphs and write down all relevant mathematics.

  1. Solve for x.

(a) sin x +

2 = − sin x in the interval [0, 2 π).

Solution:

sin x +

2 = − sin x

2 sin x = −

sin x = −

The reference is 45◦^ and the solutions occur in the third and fourth quadrant, so

x = 45

  • 180

◦ = 225

5 π

4

and x = 360

◦ − 45

◦ = 315

7 π

4

(b) 2 sin^2 x − sin x − 1 = 0 in the interval [0, 2 π).

Solution:

2 sin

2 x − sin x − 1 = 0

(2 sin x + 1) (sin x − 1) = 0.

Which gives two equations to solve.

sin x = −

and sin x = 1.

The reference for the first equation is 30◦^ and the solutions occur in the third and fourth quadrant, so

x = 30

  • 180

◦ = 210

7 π

6

and x = 360

◦ − 30

◦ = 330

11 π

6

and the solution to the second equation is x =

π

2

(^1) This document was prepared by Ron Bannon using LATEX 2 ε.