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

    $\displaystyle 2t^2 + t - 1 = 0$.

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