Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - finding solutions in trig

  1. #1
    Junior Member
    Joined
    Jun 2008
    From
    San Diego
    Posts
    26

    finding solutions in trig

    Hey i need help with this problem

    sin 3x= sqrt3/2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    You mean this, right?

    Sin\,3x = \frac{\sqrt{3}}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by NoAsherelol View Post
    Hey i need help with this problem

    sin 3x= sqrt3/2
    Where are you stuck? What have you tried?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    From
    San Diego
    Posts
    26
    Quote Originally Posted by Jonboy View Post
    You mean this, right?

    Sin\,3x = \frac{\sqrt{3}}{2}

    i meant Sin\,2x = \frac{\sqrt{3}}{2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by NoAsherelol View Post
    i meant Sin\,2x = \frac{\sqrt{3}}{2}
    The questions asked in post # 3 still apply.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by NoAsherelol View Post
    i meant Sin\,2x = \frac{\sqrt{3}}{2}
    Man, this made me really laugh!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2008
    From
    San Diego
    Posts
    26
    Quote Originally Posted by mr fantastic View Post
    The questions asked in post # 3 still apply.

    ih im sorry, this is my problem, Sin\,x = \frac{\sqrt{3}}{2} is 30 degrees i believe, im having toruble finding all the solutions to this problem
    i also tried graphing ot sinf all of the solutions.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by NoAsherelol View Post
    ih im sorry, this is my problem, Sin\,x = \frac{\sqrt{3}}{2} is 30 degrees i believe, im having toruble finding all the solutions to this problem
    i also tried graphing ot sinf all of the solutions.
    We've gone from \sin (3x) = \frac{\sqrt{3}}{2} to \sin (2x) = \frac{\sqrt{3}}{2} to \sin (x) = \frac{\sqrt{3}}{2}.

    Which one is it?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jun 2008
    From
    San Diego
    Posts
    26
    Quote Originally Posted by mr fantastic View Post
    We've gone from \sin (3x) = \frac{\sqrt{3}}{2} to \sin (2x) = \frac{\sqrt{3}}{2} to \sin (x) = \frac{\sqrt{3}}{2}.

    Which one is it?
    its this \sin (2x) = \frac{\sqrt{3}}{2}, listen man i am very stressed out, im sorry about the crazy post
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NoAsherelol View Post
    its this \sin (2x) = \frac{\sqrt{3}}{2}, listen man i am very stressed out, im sorry about the crazy post
    first thing you should know, is that if \sin x = \frac {\sqrt{3}}2 then x = 60 degrees or x =  120 degrees. that is only for 0 \le x \le 360. for all solutions you need something different. you need to tell us what solutions you are looking for. and furthermore, are you sure you shouldn't be working in radians, as opposed to degrees?

    \sin 2x = \frac {\sqrt{3}}2

    \Rightarrow 2x = 60^o \mbox{ or } 120^o for 0^o \le x \le 360^o

    \Rightarrow x = 30^o \mbox{ or } x = 60^o for 0^o \le x \le 360^o
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by NoAsherelol View Post
    ih im sorry, this is my problem, Sin\,x = \frac{\sqrt{3}}{2} is 30 degrees i believe, im having toruble finding all the solutions to this problem
    i also tried graphing ot sinf all of the solutions.
    mr fantastic got confused also, but he did not laugh. He has strong self-discipline.

    Sorry if I laughed when you first wrote sin(3x) when you actually meant sin(2x).

    So it is really for all the values of x when sin(x) = [sqrt(3)]/2.

    sin(60degrees) = [sqrt(3)]/2. Not 30 degrees.

    In one cycle, or in one revolution in the unit circle, or from 0 degrees up to 360 degees, there are only two quadrants where the sine value is positive. It is only in the 1st and 2nd quadrants.
    In the 1st quadrant, the angle, or x here, is 60 degrees.
    In the 2nd quadrant, x = 180 -60 = 120 degrees.

    In the next revolution, or from 360deg up to 720 deg, there again only 2 possible x's.
    In the 1st quadrant, x = 360 +60 = 420 deg
    In the 2nd quadrant, x = 360 +120 = 480 deg.

    In the next revolution, or from 720deg up to 1080deg,
    1st uqdrant, x = 60 +720 = 780deg
    2nd quadrant, x = 120 +720 = 840 deg.

    Etc.

    So for all the values of x in all revolutions,
    x = 60 +n(360), and x = 120 +n(360), in degrees --------answer
    where n = any non-negative integer.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    memorize this table (i put the radian measures over the degree measures):

    \begin{array}{c|c|c|c}  & \frac {\pi}6 & \frac {\pi}4 & \frac {\pi}3  \\ \theta & & & \\ & 30^o & 45^o & 60^o  \\ \hline & & & \\ \sin \theta & \frac 12 & \frac {\sqrt {2}}2 & \frac {\sqrt {3}}2 \\ & & & \\ \hline & & & \\ \cos \theta & \frac {\sqrt{3}}2 & \frac {\sqrt{2}}2 & \frac 12 \\ & & &  \end{array}

    that table tells you the sine and cosine of the special angles 30, 45 and 60 degrees (and note that, if you know the sine and cosine for an angle, you know the tangent of the angle as well. since tan(x) = sin(x)/cos(x)). all other special angles you can know from the graphs, which you should know anyway. your knowledge of reference angles allow you to deal with the analogous angles in the other quadrants (see ticbol's post)

    here's the mnemonic for filling out the table. fill out the theta, sine and cosine columns. allow the first row, count up with the angles, 30, 45, 60. then, in the sine row, count up from 1 to 3, go the opposite way with cosine. then, divide everything by 2, then square root all the numerators, and you get that table
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Jun 2008
    From
    San Diego
    Posts
    26
    Quote Originally Posted by Jhevon View Post
    first thing you should know, is that if \sin x = \frac {\sqrt{3}}2 then x = 60 degrees or x =  120 degrees. that is only for 0 \le x \le 360. for all solutions you need something different. you need to tell us what solutions you are looking for. and furthermore, are you sure you shouldn't be working in radians, as opposed to degrees?

    \sin 2x = \frac {\sqrt{3}}2

    \Rightarrow 2x = 60^o \mbox{ or } 120^o for 0^o \le x \le 360^o

    \Rightarrow x = 30^o \mbox{ or } x = 60^o for 0^o \le x \le 360^o
    as far as intervals go i dont know, it just asked me to find all solutions, and yes i think its in radians thanx
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NoAsherelol View Post
    as far as intervals go i dont know, it just asked me to find all solutions
    that's fine. see ticbol's solution. he gives you all (that's what the n(360) is for).
    and yes i think its in radians thanx
    this, however, is something you should have said earlier, especially since you mentioned degrees and misled everyone. no worries though. just convert ticbol's solution to radians.

    we go from degrees to radians by multiplying by \frac {\pi}{180}
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by Jhevon View Post
    here's the mnemonic for filling out the table. fill out the theta, sine and cosine columns. allow the first row, count up with the angles, 30, 45, 60. then, in the sine row, count up from 1 to 3, go the opposite way with cosine. then, divide everything by 2, then square root all the numerators, and you get that table
    I hadn't seen that mnemonic before. Thanks!

    Quote Originally Posted by NoAsherelol View Post
    as far as intervals go i dont know, it just asked me to find all solutions, and yes i think its in radians thanx
    In this case we are looking for an angle (because the trigonometric functions are functions of angles), and since any given angle can be represented in infinitely many ways, the equation has infinitely many solutions.

    One way to express the general solution set is to introduce an integer n as a parameter:

    We have \sin2x = \frac{\sqrt3}2.

    You should know (if not, listen to Jhevon!) that \sin{\frac\pi3} = \sin60^\circ = \frac{\sqrt3}2. This means that 60 is one possible choice for the sought angle. So we have

    2x = 60^\circ\Rightarrow x = 30^\circ

    for quadrant I. If you look at the unit circle, you should see that there is another angle in quadrant II which, using a reference angle, you can determine as 180^\circ - 60^\circ = 120^\circ, so

    2x = 120^\circ\Rightarrow x = 60^\circ

    We have two choices for x here. But we can add any multiple of 360 to an angle's measure without changing the angle it represents. So the general solution is

    2x = 60^\circ + 360^\circ n,\text{ or }2x = 120^\circ + 360^\circ n

    \Rightarrow x = 30^\circ + 180^\circ n,\text{ or }x = 60^\circ + 180^\circ n,\;n\in\mathbb{Z}

    If you aren't familiar, the notation n\in\mathbb{Z} just means that n is an integer.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Finding Exact Solutions Trig Equations
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: December 11th 2011, 04:42 PM
  2. Finding all solutions to Trig Equations
    Posted in the Trigonometry Forum
    Replies: 10
    Last Post: April 18th 2011, 01:43 PM
  3. Finding all solutions to the trig equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 11th 2010, 10:00 PM
  4. trig solutions
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 24th 2009, 11:54 AM
  5. Trig question finding all solutions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 27th 2006, 04:52 PM

Search Tags


/mathhelpforum @mathhelpforum