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Math Help - Trigonometric Equations

  1. #1
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    Trigonometric Equations

    Solve each trigonometric equation on the interval
    0 <or=to x < 2pi....Does this mean to find all the answers between 0 degrees and 360 degrees?

    (1) cos^2 (x) - sin^2 (x) + sin (x) = 0

    My attempt:

    cos^2 (x) = 1 - sin^2 (x)

    I replace cos^2 (x) with 1 - sin^2 (x) in the equation, right?

    I get this:

    1 - sin^2 (x) - sin^2 (x) + sin (x) = 0

    Do I factor out sin(x)? If so, what happens next to get me on the right track to the answer? OR....do I rearrange the sin(x) terms to be in the form ax^2 + bx + c?

    (2) sin(4x) - sin(6x) = 0

    I know there is a sin(2x) but I have never seen a 4x or 6x applied to a trig function.

    Does sin(4x) = sin(2x) + sin(2x)?

    Does sin(6x) = sin(3x) + sin(3x)?

    If so, where do I go from there?


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  2. #2
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    Quote Originally Posted by magentarita View Post
    Solve each trigonometric equation on the interval
    0 <or=to x < 2pi....Does this mean to find all the answers between 0 degrees and 360 degrees?

    (1) cos^2 (x) - sin^2 (x) + sin (x) = 0

    My attempt:

    cos^2 (x) = 1 - sin^2 (x)

    I replace cos^2 (x) with 1 - sin^2 (x) in the equation, right?

    I get this:

    1 - sin^2 (x) - sin^2 (x) + sin (x) = 0

    Do I factor out sin(x)? If so, what happens next to get me on the right track to the answer? OR....do I rearrange the sin(x) terms to be in the form ax^2 + bx + c?



    [snip]
    1 - \sin^2 x - \sin^2 x + \sin x = 0

    \Rightarrow 2 \sin^2 x - \sin x - 1 = 0

    \Rightarrow (2 \sin x + 1)(- \sin x + 1) = 0

    etc.
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  3. #3
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    Quote Originally Posted by magentarita View Post
    [snip]
    (2) sin(4x) - sin(6x) = 0

    I know there is a sin(2x) but I have never seen a 4x or 6x applied to a trig function.

    Does sin(4x) = sin(2x) + sin(2x)?

    Does sin(6x) = sin(3x) + sin(3x)?

    If so, where do I go from there?

    General solutions:

    \sin (4x) = \sin (6x)

    Case 1: 4x = 6x + 2 n \pi where n is an integer.

    Case 2: 4x = (\pi - 6x) + 2 n \pi where n is an integer.

    In each case use the general solution to solve for x over the given domain.
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  4. #4
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    Hello, magentarita!

    Solve on the interval  0 \leq x < 2\pi

    (2)\;\;\sin(4x) - \sin(6x) \:= \:0


    Does \sin(4x) \:= \:\sin(2x) + \sin(2x) ? . . . . sorry, no
    Sum-to-Product identity: . \sin A - \sin B \;=\;2\cdot\cos\!\left(\frac{A+B}{2}\right)\cdot\s  in\!\left(\frac{A-B}{2}\right)


    We have: . \sin(4x) - \sin(6x) \;=\;0

    . . 2\cdot\cos\!\left(\frac{4x+6x}{2}\right)\cdot\sin\  !\left(\frac{4x-6x}{2}\right)\;=\;0

    . . . . . . 2\cdot\cos(5x)\cdot\sin(\text{-}x) \;=\;0

    . . . . . . \text{-}2\cdot\cos(5x)\cdot\sin(x) \;=\;0


    We have: . \sin(x) \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:0,\:\pi}


    And: . \cos(5x) \:=\:0\quad\Rightarrow\quad 5x \:=\:\frac{\pi}{2} + \pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{10} + \frac{\pi}{5}n

    . . Hence: . \boxed{x \;=\;\frac{\pi}{10},\;\frac{3\pi}{10},\;\frac{\pi}  {2},\;\frac{7\pi}{10},\;\frac{9\pi}{10},\;\frac{11  \pi}{10},\;\frac{13\pi}{10},\;\frac{3\pi}{2},\;\fr  ac{17\pi}{10},\:\frac{19\pi}{10}}



    This results in the same answers as in Mr. F's excellent/elegant solution.
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  5. #5
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    Smile Thanks

    I thank you both, especially Soroban for making life simple for students like me.
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