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Math Help - Solving Trigonemtric Equations

  1. #1
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    Solving Trigonemtric Equations

    2(tan x) + sec x = 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quan View Post
    2(tan x) + sec x = 1
    2 \tan x + \sec x = 1

    \Rightarrow \frac {2 \sin x}{\cos x} + \frac 1{\cos x } - 1 = 0

    \Rightarrow \frac {2 \sin x + 1 - \cos x}{\cos x} = 0

    can you take it from here?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    2 \tan x + \sec x = 1

    \Rightarrow \frac {2 \sin x}{\cos x} + \frac 1{\cos x } - 1 = 0

    \Rightarrow \frac {2 \sin x + 1 - \cos x}{\cos x} = 0

    can you take it from here?
    so i got (2sinX + (sin^2 x)) / cos x

    i can't seem to convert them into other identities
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quan View Post
    so i got (2sinX + (sin^2 x)) / cos x

    i can't seem to convert them into other identities
    no.

    1 - \cos^2 x = \sin^2 x

    you have 1 - \cos x here, you cannot apply that identity.

    we have: \frac {2 \sin x + 1 - \cos x }{\cos x } = 0

    \Rightarrow 2 \sin x + 1 - \cos x = 0 .....(a fraction is zero when the numerator is zero, provided the denominator is not zero at the same time)

    \Rightarrow 2 \sin x + 1 = \cos x

    \Rightarrow \left( 2 \sin x + 1 \right)^2 = \cos^2 x

    \Rightarrow \left( 2 \sin x + 1 \right)^2 = 1 - \sin^2 x

    now can you continue?
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    no.

    1 - \cos^2 x = \sin^2 x

    you have 1 - \cos x here, you cannot apply that identity.

    we have: \frac {2 \sin x + 1 - \cos x }{\cos x } = 0

    \Rightarrow 2 \sin x + 1 - \cos x = 0 .....(a fraction is zero when the numerator is zero, provided the denominator is not zero at the same time)

    \Rightarrow 2 \sin x + 1 = \cos x

    \Rightarrow \left( 2 \sin x + 1 \right)^2 = \cos^2 x

    \Rightarrow \left( 2 \sin x + 1 \right)^2 = 1 - \sin^2 x

    now can you continue?
    Thanks I got it, however I want to know how you got the ^2 on both sides
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quan View Post
    Thanks I got it, however I want to know how you got the ^2 on both sides
    i just squared both sides. the result is a quadratic in sin(x)
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