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Thread: trigonometry

  1. #1
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    trigonometry

    solve for x
    2 Sin^2X + 2 SinX CosX - 1= 0
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  2. #2
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    Dear william,

    use the following identity:

    $\displaystyle 2sin(x)cos(x) = sin(2x)$
    $\displaystyle 1 = cos^2(x) + sin^2(x)$
    $\displaystyle cos^2(x) - sin^2(x) = cos(2x)$
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  3. #3
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    Hello, william!

    You should be familiar with these identities:

    . . $\displaystyle \sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 2\sin^2\!\theta \:=\:1 - \cos2\theta$

    . . $\displaystyle 2\sin\theta\cos\theta \:=\:\sin2\theta$


    Solve for $\displaystyle x\!:\;\;2\sin^2\!x + 2\sin x\cos x - 1\:=\: 0$

    We have: .$\displaystyle \underbrace{2\sin^2\!x} + \underbrace{2\sin x\cos x} -1 \;=\;0$

    . . . . . . . $\displaystyle \overbrace{1-\cos2x} + \overbrace{\sin2x} - 1 \;=\;0$


    . . $\displaystyle \sin2x \:=\:\cos 2x \quad\Rightarrow\quad \frac{\sin2x}{\cos2x} \:=\:1 \quad\Rightarrow\quad \tan2x \:=\:1$


    Therefore: .$\displaystyle 2x \:=\:\frac{\pi}{4} + \pi n \quad\Rightarrow\quad x \;=\;\frac{\pi}{8} + \frac{\pi}{2}n\;\;\text{ for any integer }n$

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  4. #4
    Senior Member euclid2's Avatar
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    Quote Originally Posted by william View Post
    solve for x
    2 Sin^2X + 2 SinX CosX - 1= 0
    sin 2x = 2 sin x cos x
    and
    cos 2x = cos2x - sin2x = 1 - 2 sin2x
    so
    2 sin2x + 2 sin x cos x -1 = 0
    is equivalent to
    sin 2x - cos 2x = 0
    which is
    sin 2x = cos 2x
    or
    tan 2x = 1.
    therefore, in the first quadrant, 2x is 45o
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