Trig Solving eqns: 2 sin x + 1 = 0, cos x (2 sin x - 1) = 0,

L

Louise Johnson

Junior Member
Joined
Jan 21, 2007
Messages
103
Hello, I have four questions I have worked on for some time and would like to know how I am doing. It just takes forever with my distance education.
Any help is really appreciated and really helps me focus on what I need practice with.
Thank you everyone

Solve each equation for x, where x is greater than 0 and less than 2 pi

#1
2 sin x+1 =0 Answer 210° and 330°

#2
cos x (2 sin x-1)=0
Answer Sin x = 30° and 150°

#3 2 cos 2x=1
Answer cos x = 1/4
This one in particular I struggled with

#4 tan^2x = 1
Answer Tan x = +or- 1
 
1)
correct
sin x =-1/2
x=-30 or 210 degrees 3rd and 4th quadrant

2)
cosx [2 sinx -1]=0
cosx=0 or 2sinx-1=0
cosx=0
x= 90 or 270 degrees

2sinx -1=0
sinx=1/2
x=30 or 150 degrees
3)
cos 2x=1/2
2x= +/-60 degrees
x=+/- 30 degrees

4)
tan^2 x=1
tan x=+/-1
tanx= +1
x= 45 or 225 degrees answer
tan x=-1
x= 315 or 135 degrees
Arthur
 
First off, let me comment that your angle solutions for x should be in radians, not degrees. You should also not have ant negative angle values as solutions. The instructions clearly state that x is between 0 and 2pi.

here is the correct solution for #3

2cos(2x) = 1

cos(2x) = 1/2

since 0 < x < 2pi, 0 < 2x < 4pi

2x = pi/3, 5pi/3, 7pi/3, 11pi/3

x = pi/6, 5pi/6, 7pi/6, 11pi/6
 
Dear Skeeter
Thank you for going over these. I have been spending some time working over these and I can see what you mentioned made sense. In question three it may seem silly but where do the extra two radians come from?

2x = 7pi/3, 11pi/3 and x= 7pi/6, 11pi/6. Are they angles outside the
interval 0 < x < 2pi ?

Thanks again for your help
Louise
 
2x = 7pi/3, 11pi/3 and x= 7pi/6, 11pi/6. Are they angles outside the
interval 0 < x < 2pi ?
Click to expand...

2pi = 12pi/6 ... so, you tell me if they are "outside" the interval.
 
Hello Skeeter sorry I have taken so long however I have been out most of the day and have to go out again. (use every possible spare moment I have) I think I may have started to make some sense out of this problem. The interval of the problem is X is greater than zero but less than 360.

2x = pi/3, 5pi/3, 7pi/3, 11pi/3 these are all equal to 60

x = pi/6, 5pi/6, 7pi/6, 11pi/6 these are all equal to 30

So as long as the radians are in quadrants one and four because that is where cosine is positive then the angles are not outside the interval..

Wow I hope I am making some sense?
Thank you
Louise
 
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