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Math Help - problem set help

  1. #1
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    problem set help

    solve for 0 < x < 2PI:
    a) 2tan^2(x) - 3tan(x) - 4 = 0
    b) sin(2x) + sin(4x) = 0
    c) tan(2x) + 2cos(x) = 0

    prove the identity:
    a) sin(4x) + sin(2x) = 2sin(3x)*cos(x)
    b) tan(3x) = (3tan(x) - tan^3(x)) / (1 - 3tan^2(x))

    thank you for the help
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by lazyboy292 View Post
    s

    prove the identity:
    a) sin(4x) + sin(2x) = 2sin(3x)*cos(x)
    b) tan(3x) = (3tan(x) - tan^3(x)) / (1 - 3tan^2(x))
    a) Use \sin a+\sin b=2\sin\frac{a+b}{2}\cos\frac{a-b}{2}

    b) Use \tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}
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  3. #3
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    Hello, lazyboy292!

    Solve for 0 \leq x \leq 2\pi

    . . c)\;\;\tan(2x) + 2\cos(x) \:=\: 0

    We have: . \frac{\sin2x}{\cos2x} + 2\cos x \:=\:0

    . . . . \frac{2\sin x\cos x}{2\cos^2x-1} + 2\cos x \:=\:0

    . . . . 2\sin x\cos x + 4\cos^3\!x - 2\cos x \:=\:0


    Factor: . 2\cos x\bigg(\sin x + 2\cos^2\!x -1\bigg) \:=\:0

    . . . . 2\cos x\bigg(\sin x + 2[1-\sin^2\!x] - 1\bigg)\:=\:0

    . . . . -2\cos x\bigg(2\sin^2\!x - \sin x - 1\bigg) \:=\:0

    . . . . -2\cos x(\sin x - 1)(2\sin x + 1) \:=\:0


    And we have three equations to solve . . .

    . . -2\cos x \:=\:0\quad\Rightarrow\quad \cos x \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}

    . . \sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2}}

    . . 2\sin x + 1\:=\:0\quad\Rightarrow\quad \sin x \:=\:-\tfrac{1}{2}\quad\Rightarrow\quad\boxed{ x \:=\:\frac{7\pi}{6},\:\frac{11\pi}{6}}

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