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Math Help - Trigonometric equation

  1. #1
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    Trigonometric equation

    if \sin x+\tan x=\frac{5}{6} then \sin x\tan x=
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  2. #2
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    Re: Trigonometric equation

    What steps have you taken to try and solve this ?
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  3. #3
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    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    if \sin x+\tan x=\frac{5}{6} then \sin x\tan x=
    Dear jacks,

    \sin x+\tan x=\frac{5}{6}

    \sin x\left(1+\frac{1}{\cos x}\right)=\frac{5}{6}

    \pm\sqrt{1-\cos^{2}x}\left(\frac{1+\cos x}{\cos x}\right)=\frac{5}{6}

    (1-\cos^{2}x)\left(\frac{1+\cos x}{\cos x}\right)^2=\frac{25}{36}

    1-2\cos x-\frac{25}{36}\cos^{2}x+2\cos^{3}x-\cos^{4}x=0

    By solving this quartic equation you could obtain,

    \cos x=\begin{cases}~~0.91710326\\-0.50822898\\-1.2044371+0.83354720i\\-1.2044371-0.83354720i\end{cases}

    \sin^{2}x+\tan^{2}x=(1-\cos^{2}x)+\left(\frac{1}{\cos^{2}x}-1\right)=\left\{\begin{array}{l}~~0.347871430\\~~3  .61322011\\-0.59165723+2.44412431i\\-0.59165723-2.44412431i\end{array}\right.

    Also note that,

    (\sin x+\tan x)^2=\frac{25}{36}

    \sin^{2}x+\tan^{2}x+2\sin x\tan x=\frac{25}{36}

    \sin x\tan x=\frac{25}{72}-\frac{\left(\sin^{2}x+\tan^{2}x\right)}{2}

    \sin x\tan x=\begin{cases}~~0.17328651\\-1.45938783\\~~0.64305084-1.22206216i\\~~0.64305084+1.22206216i\end{cases}
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