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Thread: Solving trig equations

  1. #1
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    Solving trig equations

    Ok Well, I need a lot of help with this. I hav eso many problems to do and I do not even know where to start.

    I just want to get help with a few and then I'll see if I can figure out how to do them on my own..

    Ok so here is the first type of problem I have.

    Solve each trig equation.

    First off I'm not entirely sure what I'm trying to find in the following. Even then I don't really know where to start.
    1. sin x = -sqrt2/2

    2. Cos 2x=-sqrt2/2

    3. tan .5x= sqrt3

    4. sqrt3 sec x +2 =0

    I think if i know how to do those I will be able to figure out the rest of those type of problems.

    Here id the other type of problem i have to do.

    Solve the each trig equation

    1.4 sin^2 x-3=0

    For this one would I just add 3 then divide by 4. Then take the inverse sin of sqrt of 3/4 and the answer would be 1.047. If that is right then that would be one of the solutions but Im not sure how to find the other one

    2. cos x +2= 3cos x

    3. 2 sin ^2 x-3=0
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mtt0216 View Post
    Ok Well, I need a lot of help with this. I hav eso many problems to do and I do not even know where to start.

    I just want to get help with a few and then I'll see if I can figure out how to do them on my own..

    Ok so here is the first type of problem I have.

    Solve each trig equation.

    First off I'm not entirely sure what I'm trying to find in the following. Even then I don't really know where to start.
    1. sin x = -sqrt2/2

    2. Cos 2x=-sqrt2/2

    3. tan .5x= sqrt3

    4. sqrt3 sec x +2 =0

    I think if i know how to do those I will be able to figure out the rest of those type of problems.

    Here id the other type of problem i have to do.

    Solve the each trig equation

    1.4 sin^2 x-3=0

    For this one would I just add 3 then divide by 4. Then take the inverse sin of sqrt of 3/4 and the answer would be 1.047. If that is right then that would be one of the solutions but Im not sure how to find the other one

    2. cos x +2= 3cos x

    3. 2 sin ^2 x-3=0
    A) For the first lot you can take the inverse function.

    For $\displaystyle \sin(x) = -\frac{\sqrt2}{2}$

    $\displaystyle x = \arcsin \left(-\frac{\sqrt2}{2}\right)$

    You should also know your special triangles

    B) Divide by 4. Then use the difference of two squares

    $\displaystyle \left(\sin (x) - \frac{\sqrt3}{2}\right) \left(\sin (x) + \frac{\sqrt3}{2}\right)=0$

    Which will give two solutions
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  3. #3
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    Well I'm still confused. in my class we have never used arcsin so I don't even know what that is or how to use it.

    And for the second part I'm not exactly sure what you did.
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  4. #4
    Super Member bigwave's Avatar
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    2

    arcsin basically means $\displaystyle sin^{-1}\theta$ which is an inverse function

    #2

    $\displaystyle \cos{(2x)}=-\frac{\sqrt{2}}{2}$
    $\displaystyle
    \cos^{-1}\left({-\frac{\sqrt{2}}{2}}\right) = 2x$

    $\displaystyle \frac{3\pi}{4} = 2x$

    $\displaystyle \frac{3\pi}{8} = x$
    Last edited by bigwave; Mar 4th 2010 at 12:36 PM. Reason: more into
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  5. #5
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    ..........
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  6. #6
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    thanks for the help guys.

    I did some of the problems and I wanted to see if i got the right answer for them.

    so

    $\displaystyle sin=-\sqrt{2} /2$

    x= $\displaystyle \frac{5pi}{4} $ or $\displaystyle \frac{7pi}{4} $

    ==========================
    cos 2x= $\displaystyle \frac{\sqrt{2}}{2}$

    x= $\displaystyle \frac{3pi}{8} $ or $\displaystyle \frac{5pi}{8} $

    ===========================

    $\displaystyle sin(x+\frac{pi}{4} )$=$\displaystyle \frac{-\sqrt{2} }{2} $

    x= $\displaystyle 6\frac{pi}{4} $ or $\displaystyle 4\frac{pi}{4} $

    =========================
    $\displaystyle
    tan.5x=\sqrt{3} $

    $\displaystyle x= \frac{pi}{1.5} $
    =========================
    $\displaystyle 2sinx+\sqrt{3}=0 $

    $\displaystyle x= 5\frac{pi}{3} $ or $\displaystyle 4\frac{pi}{3} $

    =============================

    $\displaystyle 4 tanx-2=2$

    $\displaystyle x= \frac{pi}{4} $





    I wasn't able to get these 2.
    $\displaystyle tan(x-30deg)=\sqrt{3} $

    $\displaystyle \sqrt{3} secx+2=0$



    Also how do I show that an equation has an infinite amount of solutions
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