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Math Help - Solve for x: cos x + cos 3x = sin x + sin 3x

  1. #1
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    Solve for x: cos x + cos 3x = sin x + sin 3x

    I copied down the solutions from my teacher for this question and i understand up to this point:

    ⇒ 0 = sin 6x - sin 2x

    The next step it is:

    ⇒ 0 = 2cos ((6x+2x)/2) sin ((6x-2x)/2)

    I don't understand how that the addition formulas were used to get to this step.


    The solution i copied down is below:

    cos x + cos 3x = sin x + sin 3x
    ⇒cos x - sin x = sin 3x - cos 3x
    now, squaring on both sides,
    (cos x - sin x)^2 = (sin 3x - cos 3x)^2
    ⇒(cos x)^2 2 sin x cos x + (sin x)^2 = (sin 3x)^2 - 2sin 3x cos 3x+ (cos 3x)^2
    ⇒ 1-2 sin x cos x = 1-2sin 3x cos 3x
    ⇒ -2 sin x cos x = -2sin 3x cos 3x
    ⇒ sin x cos x = sin 3x cos 3x
    multiplying both sides by 2,
    ⇒ 2sin x cos x = 2sin 3x cos 3x
    ⇒ sin 2x = sin 2(3x)
    ⇒ sin 2x = sin 6x
    ⇒ 0 = sin 6x - sin 2x
    ⇒ 0 = 2cos ((6x+2x)/2) sin ((6x-2x)/2)
    ⇒ 0 = 2cos 4x sin 2x
    ⇒ 0 = cos 4x sin 2x
    case 1:
    ⇒ cos 4x = 0
    ⇒ cos 4x = cos π/2
    ⇒ 4x = 2nπ + π/2
    ⇒ x = nπ/2 + π/8
    case 2:
    ⇒ sin 2x = 0
    ⇒ sin 2x = sin π
    ⇒ 2x = nπ + ((-1)^n)(π)
    ⇒ x = nπ/2 + ((-1)^n)(π)/2

    ⇒ x = [nπ/2 + π/8] or [nπ/2 + ((-1)^n)(π)/2]
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  2. #2
    Senior Member pacman's Avatar
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    see this graph, only 6 solutions. figure out why?

    Attached Thumbnails Attached Thumbnails Solve for x: cos x + cos 3x = sin x + sin 3x-asd.gif  
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  3. #3
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    Hello, BuffaloSoulja!

    There are four sum-to-product identities:

    . . \begin{array}{cccc}(1) & \sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\[-3mm]<br />
(2) & \sin A - \sin B &=& 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\[-3mm]<br />
(3) & \cos A + \cos B &=&2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{  A-B}{2}\right) \\ \\[-3mm]<br />
(4) & \cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)<br /> <br />
\end{array}


    Your teacher's solution is rather silly . . .

    He/she used (2) in that mysterious step.

    What makes it silly is that he/she could have used these identities first
    . . and save a lot of work.


    \cos x + \cos 3x \:=\: \sin x + \sin 3x
    We have: . \cos3x + \cos x \:=\:\sin3x + \sin x

    Using (1) and (3), we have: . 2\cos2x\cos x \;=\;2\sin2x\cos x \quad\Rightarrow\quad 2\cos2x\cos x - 2\sin2x\sin x \;=\;0

    Factor: . 2\cos x(\cos2x - \sin 2x) \;=\;0


    And we have two equations to solve:

    2\cos x \:=\:0 \quad\Rightarrow\quad \cos x \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2} + \pi n}

    \cos2x\;-\;\sin2x\:=\:0 \quad\Rightarrow\quad \sin2x\:=\:\cos2x \quad\Rightarrow\quad \frac{\sin2x}{\cos2x} \:=\:1 \quad\Rightarrow\quad \tan2x \:=\:1

    . . 2x \:=\:\frac{\pi}{4} + \pi n \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{8} + \frac{\pi}{2}n}

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