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Math Help - can't solve a trigo equ

  1. #1
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    can't solve a trigo equ

    i can't solve this trigo equ.
     2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0
     sin 2xcos^2 x - (1+sinx )(cos2x)= 0
     tan2x = \frac {1+sinx}{cos^2x}
    ?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
     2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0
    I used the identities 2\sin x\cos x=\sin 2x and \cos^2x-\sin^2x=\cos 2x
    Now, the equation becomes
    \sin 2x\cos x-\cos 2x-\sin x\cos 2x=0\Leftrightarrow \sin x-\cos 2x=0
    (I used the formula \sin(a-b)=\sin a\cos b-\sin b\cos a)
    Using the formula \cos 2x=1-2\sin^2x the equation becomes 2\sin^2x+\sin x-1=0
    Let \sin x=t and the equation 2t^2+t-1=0 has the roots t_1=-1, \ t_2=\frac{1}{2}
    So you have to solve the equations \sin x=-1 and \sin x=\frac{1}{2}
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  3. #3
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    Quote Originally Posted by afeasfaerw23231233 View Post
    i can't solve this trigo equ.
     2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0
     sin 2xcos^2 x - (1+sinx )(cos2x)= 0
     tan2x = \frac {1+sinx}{cos^2x}
    ?
    Go back to line 1. Substitute \cos^2 x = 1 - \sin^2 x and expand. Let t = \sin x. You end up with (but you better check):

    2t^2 + t - 1 = 0.

    Solve for t. Then solve t = \sin x for x.
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