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Math Help - Various trigonometric equations

  1. #1
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    Various trigonometric equations

    Hello, I have this worksheet and there are a few problems I need clarification on. I've done most of them but my answers all turn out weird.

    If cos(x) = -7/9, tan(x) < 0, find cos2(x)

    Find all the solutions of sin5x + sinx = sin3x in the interval [0, 2pi)

    Rewrite sin3xcos4y as a sum

    Find all the solutions of 3cos(x/4) in the interval of [0, 2pi)


    Thanks
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  2. #2
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    Hello, littlelisa!

    If \cos x = \text{-}\frac{7}{9},\;\;\tan x  < 0, find \cos2x
    Cosine is negative in Quadrants 2 and 3.
    Tangent is negative in Quadrants 2 and 4.
    . . Hence, x is in Quadrant 2.

    Identity: . \cos2\theta \:=\:2\cos^2\!\theta -1

    So we have: . \cos2x \;=\;2\left(\text{-}\frac{7}{9}\right)^2 - 1 \;=\;\frac{17}{81}



    Find all the solutions of \sin5x + \sin x \:=\: \sin3x in the interval [0, 2\pi)
    Sum-to-product identity: . \sin A + \sin B \;=\;2\sin\left(\frac{A+B}{2}\right)\cos\left(\fra  c{A-B}{2}\right)

    So: . \sin5x+\sin x \:=\:2\sin\left(\frac{5x+x}{2}\right)\cos\left(\fr  ac{5x-x}{2}\right) \;=\;2\sin3x\cos2x

    The equation becomes: . 2\sin3x\cos2x\:=\:\sin3x \quad\Rightarrow\quad 2\sin3x\cos2x - \sin3x \:=\:0

    . . Factor: . \sin3x(2\cos2x - 1) \:=\:0


    We have two equations to solve . . .

    \sin3x \:=\:0\quad\Rightarrow\quad 3x \:=\:0, \pi, 2\pi, 3\pi, 4\pi, 5\pi
    . . Therefore: . \boxed{x \:=\:0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}}

    2\cos2x-1\:=\:0\quad\Rightarrow\quad \cos2x \:=\:\frac{1}{2}\quad\Rightarrow\quad 2x \:=\:\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3},\  frac{11\pi}{3}
    . . Therefore: . \boxed{x \;=\;\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}}



    Rewrite \sin3x\cos4y as a sum.
    Product-to-sum identity: . \sin A\cos B \;=\;\frac{1}{2}\bigg[\sin(A-B) + \sin(A + B)\bigg]

    Therefore: . \sin3x\cos4y \;=\;\frac{1}{2}\sin(3x-4y) + \frac{1}{2}\sin(3x+4y)




    Find all the solutions of 3\cos\left(\frac{x}{4}\right) in the interval [0, 2\pi)
    You left out part of the equation . . .

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  3. #3
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    Thanks and sorry about the last equation it equals 0.

    3cos(x/4)=0

    Thanks again I really appreciate it!!!
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