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Math Help - Trigonometric equation

  1. #1
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    Trigonometric equation

    Can someone please help me solve the following trig equations please? Thanks

    1-tan^2(x)/1+tan^2(x) + cos(x) = 0

    (sqrt(3)+1)cos^2(x) - 1 = (sqrt(3)-1)sin(x)cos(x)

    Thanks heaps
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  2. #2
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    Quote Originally Posted by deltaxray View Post
    Can someone please help me solve the following trig equations please? Thanks

    1-tan^2(x)/1+tan^2(x) + cos(x) = 0

    (sqrt(3)+1)cos^2(x) - 1 = (sqrt(3)-1)sin(x)cos(x)

    Thanks heaps
    Hi

    Did u mean solve the trigonometric equation ? If so ,

    For (1) , \frac{1-\tan^2 x}{1+\tan^2 x}+\cos x=0

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

    \frac{\cos^2 x-\sin^2 x}{\cos^2 x}(\cos^2 x)+\cos x=0

    2\cos^2 x+\cos x-1=0

    Then (2\cos x-1)(\cos x +1)=0 , i think you should be able to continue from here .
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  3. #3
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    Thanks Mate.
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  4. #4
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    Hello deltaxray

    Welcome to Math Help Forum!
    Quote Originally Posted by deltaxray View Post
    Can someone please help me solve the following trig equations please? Thanks

    1-tan^2(x)/1+tan^2(x) + cos(x) = 0

    (sqrt(3)+1)cos^2(x) - 1 = (sqrt(3)-1)sin(x)cos(x)

    Thanks heaps
    For question 2, divide both sides by \cos^2x:

    \sqrt3+1 - \sec^2x=(\sqrt3-1)\frac{\sin x\cos x}{\cos^2x}

    Then express everything in terms of \tan x. Re-arrange as a quadratic in \tan x, and you'll find it factorises into:

    (\tan x -1)(\tan x + \sqrt3)=0

    Can you fill in the gaps?

    Grandad
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  5. #5
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    Thanks granddad. I got it
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  6. #6
    Senior Member pacman's Avatar
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    by inspection, pi is a root to the equation, the other 2 roots are found already; x = {pi/3, pi, 5pi/3}

    the graph below confirms it
    Attached Thumbnails Attached Thumbnails Trigonometric equation-asd.gif  
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