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Math Help - Trigonometric Equation

  1. #1
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    Trigonometric Equation

    (a) Express 3cosx + 2sinx in the form Rcos(x – α), where R and α are constants with R 1 0 and
    0 < α < 90. [3]
    (b) Find all values of x between 0 and 360 satisfying
    3cosx + 2sinx = 1. [3]

    For (a), I have \sqrt{13}\ cos\ (x - 33.7)

    and in (b), I got up to saying cos\ (x - 33.7) = \frac{1}{\sqrt1{3}}, but I have no idea what is done after this to get the values of x.

    How is this done? Thanks if you're able to help me
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by db5vry View Post
    (a) Express 3cosx + 2sinx in the form Rcos(x α), where R and α are constants with R 1 0 and
    0 < α < 90. [3]
    (b) Find all values of x between 0 and 360 satisfying
    3cosx + 2sinx = 1. [3]

    For (a), I have \sqrt{13}\ cos\ (x - 33.7)

    and in (b), I got up to saying cos\ (x - 33.7) = \frac{1}{\sqrt1{3}}, but I have no idea what is done after this to get the values of x.

    How is this done? Thanks if you're able to help me
    For b) Put the equation in terms of cosine (or sine) by making use of the pythagorean relationship \sin{x}=\sqrt{1-\cos^2{x}}.

    So we have

    3\cos{x}+2\cdot\sqrt{1-\cos^2{x}}=1

    Square both sides

    (3\cos{x}-1)^2=(-2\cdot\sqrt{1-\cos^2{x}})^2

    9\cos^2{x}-6\cos{x}+1=4-4\cos^2{x}

    Set the equation equal to zero and solve the quadratic in x.

    13\cos^2{x}-6\cos{x}-3=

    By the quadratic formula

    \cos{x}=\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}

    Therefor x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right]
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    For b) Put the equation in terms of cosine (or sine) by making use of the pythagorean relationship \sin{x}=\sqrt{1-\cos^2{x}}.

    So we have

    3\cos{x}+2\cdot\sqrt{1-\cos^2{x}}=1

    Square both sides

    (3\cos{x}-1)^2=(-2\cdot\sqrt{1-\cos^2{x}})^2

    9\cos^2{x}-6\cos{x}+1=4-4\cos^2{x}

    Set the equation equal to zero and solve the quadratic in x.

    13\cos^2{x}-6\cos{x}-3=

    By the quadratic formula

    \cos{x}=\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}

    Therefor x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right]
    I see how you do that, and I see how it is right, but that's not the way I'm meant to do it. I need to solve it from the form Rcos(x - a) = 1. That's what I would like to learn how to do because it's troubled me in the past
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