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Math Help - Questions on trigo. equation

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    Questions on trigo. equation

    Find general solution for the following
    1. \cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}
    2. \cos{(2x+60^o)}=\sin{x}

    can anyone help?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by acc100jt View Post
    Find general solution for the following
    1. \cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}
    2. \cos{(2x+60^o)}=\sin{x}

    can anyone help?
    expand using the addition and double angle formulas and then see what you can do.

    recall:

    \cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta

    \sin 2 \theta = 2 \sin \theta \cos \theta

    \cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta


    note: \cos 3 x = \cos (2x + x) and \sin 3x = \sin (2x + x)
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  3. #3
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by acc100jt View Post
    Find general solution for the following
    1. \cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}
    \cos 3x+\sqrt{3}\sin 3x=2\cos x
    Divide both members by 2:
    \displaystyle\frac{1}{2}\cos 3x+\frac{\sqrt{3}}{2}\sin 3x=\cos x
    \cos 60^{\circ}\cos 3x+\sin 60^{\circ}\sin 3x=\cos x
    \cos(60^{\circ}-3x)=\cos x
    \cos(60^{\circ}-3x)-\cos x=0
    \displaystyle 2\sin(2x-30^{\circ})\sin (30^{\circ}-x)=0

    Now, continue.
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  4. #4
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    Quote Originally Posted by acc100jt View Post
    Find general solution for the following
    1. \cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}
    2. \cos{(2x+60^o)}=\sin{x}

    can anyone help?
    Umm, these are really very hard to solve. Need tricks.

    Even when I was yet in High School, I resort to iteration, or trial and error, in solving equations that I didn't know how to solve the usual ways.
    Problem 1 is is a good candidate for my iteration.

    By luck, or because of the sqrt(3) in the equation, I tried first, x = 30degrees.
    cos(3*30deg) =? 2cos(30deg) -sqrt(3)*sin(3*30deg)
    cos(90deg) =? 2cos(30deg) -sqrt(3)*sin(90deg)
    0 =? 2[sqrt(3) / 2] -sqrt(3)*(1)
    0 =? sqrt(3) -sqrt(3)
    0 =? 0
    Yes, so x = 30deg is OK.

    I was just lucky to hit it in one try only. So, 30deg is a solution. There might be some more but I don't know how, or I don't want, to continue after that.

    ------------------------------
    For the second problem, another trick.

    cos(2x +60deg) = sin(x) ----------(ii)

    Expanding that will lead to an equation that is as complicated as the first one, so another dip into our bag of tricks:

    We know that cosA = sin(90deg -A).
    That means A +(90deg -A) = 90 deg.

    Or, cosA = sinB -----if angles A and B are complementary.

    So, using that on (ii),
    (2x +60deg) +x = 90deg
    3x = 90 -60
    x = 30/3
    x = 10 deg ------------a solution.

    Check,
    cos(2*10deg +60deg) =? sin(10deg)
    cos(80deg) =? sin(10deg)
    Using a calculator,
    0.173648178 =? 0.173648178
    Yes, so, x = 10 degrees is correct.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    \cos 3x+\sqrt{3}\sin 3x=2\cos x
    Divide both members by 2:
    \displaystyle\frac{1}{2}\cos 3x+\frac{\sqrt{3}}{2}\sin 3x=\cos x
    \cos 60^{\circ}\cos 3x+\sin 60^{\circ}\sin 3x=\cos x
    \cos(60^{\circ}-3x)=\cos x
    \cos(60^{\circ}-3x)-\cos x=0
    \displaystyle 2\sin(2x-30^{\circ})\sin (30^{\circ}-x)=0

    Now, continue.
    nice!
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