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avatar+466 

Solve, given that (0o < x < 360o): 2 sin2 x - 3 sin x + 1 = 0

I have to solve for x, and x has to be an angle. sad

 Jan 6, 2016

Best Answer 

 #11
avatar+104712 
+10

Sorry, folks.......the answer to this question is shown by the graph below :

 

 

The solutions occur at 30, 90 and 150  degrees......

 

See my "workings" here : http://web2.0calc.com/questions/unanswered-question-x-being-an-angle-solve

 

 

cool cool cool

 Jan 8, 2016
edited by CPhill  Jan 8, 2016
 #1
avatar
+5

Solve for x:
1-3 sin(x)+2 sin^2(x) = 0

The left hand side factors into a product with two terms:
(sin(x)-1) (2 sin(x)-1) = 0

Split into two equations:
sin(x)-1 = 0 or 2 sin(x)-1 = 0

Add 1 to both sides:
sin(x) = 1 or 2 sin(x)-1 = 0

Take the inverse sine of both sides:
x = pi/2+2 pi n_1  for  n_1  element Z
   or  2 sin(x)-1 = 0

Add 1 to both sides:
x = pi/2+2 pi n_1  for  n_1  element Z
   or  2 sin(x) = 1

Divide both sides by 2:
x = pi/2+2 pi n_1  for  n_1  element Z
   or  sin(x) = 1/2

Take the inverse sine of both sides:
Answer: | 
| x = pi/2+2 pi n_1  for  n_1  element Z
  or  x = (5 pi)/6+2 pi n_2  for  n_2  element Z        or x = pi/6+2 pi n_3  for  n_3  element Z

 Jan 6, 2016
 #2
avatar+466 
+5

I don't know what you tried to do, but that doesn't help me at all, sorry...

 Jan 6, 2016
 #3
avatar
0

That is because you don't understand "complex solutions"!!. If you want to go to college and study advanced Math, your better start honing up on it, because it is the only game in town at that level.

 Jan 6, 2016
 #4
avatar+466 
0

Okay, someone please get an admin in here... Guest, I know how to do math, this problem is one that I do all of the time, I just can't figure this one out. angry

 Jan 6, 2016
 #5
avatar+105509 
+5

Hi Shades :)

 

2 sin2 x - 3 sin x + 1 = 0

 

\(2 sin^2 x - 3 sin x + 1 = 0\\ let\;\; y=sin^2x\\ 2y^2-3y+1=0\\ 2y^2-2y-y+1=0\\ 2y(y-1)-1(y-1)=0\\ (2y-1)(y-1)=0\\ 2y-1=0 \qquad or \qquad y-1=0\\ 2y=1 \qquad or \qquad y=1\\ y=\frac{1}{2} \qquad or \qquad y=1\\ so\\ sin^2x=\frac{1}{2} \qquad or \qquad sin^2x=1\\ sinx=\pm \frac{1}{\sqrt{2}} \qquad or \qquad sinx=\pm1\\\)

 

\(x=45^0,135^0,225^0,315^0, 90^0,\;\;or \;\;270^0\)

.
 Jan 6, 2016
edited by Melody  Jan 7, 2016
edited by Melody  Jan 7, 2016
 #6
avatar
+5

Wake up Melody.

 Jan 6, 2016
 #7
avatar+4080 
0

HaHa, nice one guest... No offense Melody!

 Jan 6, 2016
 #8
avatar+105509 
0

Huh?  I don't do cryptic very well   laugh

 Jan 7, 2016
 #9
avatar
+5

Look again at your solution #5

 Jan 7, 2016
 #10
avatar+105509 
0

Thanks guest #9.    I have edited my answer.     laugh

 Jan 7, 2016
 #11
avatar+104712 
+10
Best Answer

Sorry, folks.......the answer to this question is shown by the graph below :

 

 

The solutions occur at 30, 90 and 150  degrees......

 

See my "workings" here : http://web2.0calc.com/questions/unanswered-question-x-being-an-angle-solve

 

 

cool cool cool

CPhill Jan 8, 2016
edited by CPhill  Jan 8, 2016

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