Results 1 to 10 of 10

Math Help - Solving for X

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    4

    Solving for X

    Alright so here is my problem, first day of college started I am in Calculus and have never took a trigonometry class in my life. Because Calculus involves some trigonometry the teacher assigned us a trig review for homework and I have no clue on how to do it, thanks for the help I get.

    Here are some of the problems:
    Provide all EXACT solutions for the variable on the interval [0,2pie)

    2sin3pie/4 = X

    2cos(x) = -1

    3sec(7pieX/3) = -2(square root of 3)

    if you could explain how u got the answers I would appreciate it thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Bran21 View Post

    2sin3pie/4 = X
    X = 2\sin\left(\frac{3\pi}{4}\right)

    \sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

    Therefore

    X = 2\frac{1}{\sqrt{2}}

    X = \frac{2}{\sqrt{2}}


    Quote Originally Posted by Bran21 View Post

    2cos(x) = -1
    2\cos(x) = -1

    \cos(x) = \frac{-1}{2}

    Cosine is negative in the 2nd and 3rd quadrants so

    x = \pi - \theta, \pi+\theta where \theta = \frac{\pi}{3}

    x = \pi - \frac{\pi}{3}, \pi+\frac{\pi}{3}

    x = \frac{2\pi}{3},\frac{4\pi}{3}
    Last edited by pickslides; August 24th 2009 at 02:16 PM. Reason: Typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    4
    Thanks so much pickslides I have a few more questions if you have the time if not youve helped me enough already and im grateful that you did.

    Cos(2Pie/3) = 2x

    0 = 1 + Sin (3x/2)

    Tanx = Cotx


    Edit: Also for the second problem how did you come up with theta = pie/3
    Also I know its Pi not Pie but Mozilla was fixing it
    Last edited by Bran21; August 24th 2009 at 02:54 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Bran21 View Post

    Edit: Also for the second problem how did you come up with theta = pie/3
    There are some special triangles you need to commit to memory if you are going to be successful in completing a course in circular trigonometry.

    Remember these Special right triangles - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2009
    Posts
    4
    Quote Originally Posted by pickslides View Post
    There are some special triangles you need to commit to memory if you are going to be successful in completing a course in circular trigonometry.

    Remember these Special right triangles - Wikipedia, the free encyclopedia
    So looking at this you decided it was 60 degrees and in radians thats pi/3 but I dont understand how you got the 60 degrees then or you looked at the -1/2 and decided from there, im sorry Im really wishing I took a trig course in high school cause its messing me up
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Assuming you are aware of the different sides on a right angled triangle then.

    \cos(\theta) = \frac{A}{H}

    Consider

    \cos(\theta) = \frac{1}{2}

    and the triangle from the wiki page above. Then the angle that yields \frac{1}{2} for cosine is \frac{\pi}{3}

    I.e \cos(\frac{\pi}{3}) = \frac{1}{2}

    Now for

    \cos(\theta) = \frac{-1}{2} You have to use the symmetry of the unit circle to find where Cosine is a negative value.

    Have a look at

    http://www.clarku.edu/~djoyce/trig/functions.html
    Last edited by pickslides; August 24th 2009 at 04:40 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2009
    Posts
    4
    Alright I think im starting to understand this, its a 30-60-90 triangle and that means there is the 60 degree angle and the cos would = 1/2 because the 1 is adjacent on the 30-60-90 and the 2 is the hypotenuse. Then you get the pi/3 from the chart.

    Alright now how would I relate tanx = cotx

    Thanks for help so far
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Bran21 View Post
    Alright I think im starting to understand this, its a 30-60-90 triangle and that means there is the 60 degree angle and the cos would = 1/2 because the 1 is adjacent on the 30-60-90 and the 2 is the hypotenuse. Then you get the pi/3 from the chart.


    Thanks for help so far
    Yep

    Quote Originally Posted by Bran21 View Post

    Alright now how would I relate tanx = cotx

    Thanks for help so far
    Consider \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}

    Now \tan(x) = \cot(x) is the same as saying

    \frac{\sin(x)}{\cos(x)} = \frac{\cos(x)}{\sin(x)}

    With a bit of manipulation you get

    \sin^2(x)= \cos^2(x)

    1-\cos^2(x)= \cos^2(x)

    1= 2\cos^2(x)

    \frac{1}{ 2} = \cos^2(x)

     \cos(x) = \frac{1}{ \sqrt{2}}

    Now go find those triangles again!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by pickslides View Post
    Yep



    Consider \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}

    Now \tan(x) = \cot(x) is the same as saying

    \frac{\sin(x)}{\cos(x)} = \frac{\cos(x)}{\sin(x)}

    With a bit of manipulation you get

    \sin^2(x)= \cos^2(x)

    1-\cos^2(x)= \cos^2(x)

    1= 2\cos^2(x)

    \frac{1}{ 2} = \cos^2(x)

     \cos(x) = {\color{red}\pm}\frac{1}{ \sqrt{2}}

    Now go find those triangles again!
    Note my correct in red. You can't neglect the fact that the value can be negative!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Thanks Chrissles.

    Always rushing to get these things out at work.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 7th 2009, 04:13 PM
  2. help solving for y
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 1st 2009, 10:25 AM
  3. help me in solving this
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 13th 2008, 01:51 PM
  4. Solving x'Px = v
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 11th 2008, 03:21 PM
  5. Replies: 3
    Last Post: October 11th 2006, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum