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Math Help - trigonometric equation

  1. #1
    Member CalcGeek31's Avatar
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    trigonometric equation

    4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
    solutions. Write answers in radians.
    4. 2sin^2(θ) =1 5. tan^2(θ) − tan(θ) = 0 7. cos^2(θ) + sin(θ) =1

    is the first and second ones just in the 45 degree family?
    Last edited by CalcGeek31; August 23rd 2008 at 10:04 AM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by CalcGeek31 View Post
    4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
    solutions. Write answers in radians.
    4. 2sin^2(θ) =1 5. tan^2(θ) − tan(θ) = 0 7. cos^2(θ) + sin(θ)

    is the first and second ones just in the 45 degree family?
    Both 1 and 2 consists of the \frac{\pi}{4} family...

    For #1, you get two angles, \vartheta=\frac{\pi}{4} and another angle. Just keep in mind the quadrants where sine is positive! I'll leave it for you to find the other angle.

    However, for #2, we see that we can manipulate the equation a little bit:

    \tan^2\vartheta-\tan\vartheta=0\implies\tan\vartheta(\tan\vartheta-1)=0

    We will get two different equations:

    \tan\vartheta=0 and \tan\vartheta=1

    Keep in mind the restriction on the angle : 0\leq\vartheta<2\pi

    Also keep in mind which quadrants tangent is positive!

    I hope this helps.

    --Chris
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  3. #3
    Member CalcGeek31's Avatar
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    what about the third one? it is = 1 will go edit that
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by CalcGeek31 View Post
    4-6 Solve the equation for θ (0 ≤θ < 2π) . Be careful. There may be several
    solutions. Write answers in radians.
    4. 2sin^2(θ) =1 5. tan^2(θ) − tan(θ) = 0 7. cos^2(θ) + sin(θ) =1

    is the first and second ones just in the 45 degree family?
    #3 : \cos^2\vartheta+\sin\vartheta=1

    First convert the eqaution into sine. Use the fact that \cos^2\vartheta=1-\sin^2\vartheta

    Thus, our equation then becomes 1-\sin^2\vartheta+\sin\vartheta=1\implies \sin^2\vartheta-\sin\vartheta=0\implies \sin\vartheta(\sin\vartheta-1)=0

    Again, we get two equations:

    \sin\vartheta=0 and \sin\vartheta=1

    Keep in mind the restriction on the angle : 0\leq\vartheta<2\pi

    I hope this helps.

    --Chris
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  5. #5
    Member CalcGeek31's Avatar
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    for #1... I just re did it 4 times and got 4 answers not 2 can anyone confirm?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by CalcGeek31 View Post
    for #1... I just re did it 4 times and got 4 answers not 2 can anyone confirm?
    Erm...

    I overlooked something pretty simple and thought of one equation, but there were two sets of equations:

    \sin\vartheta=\tfrac{1}{\sqrt{2}} and \sin\vartheta=-\tfrac{1}{\sqrt{2}}

    So you should get four solutions, not two. Sorry about that! It was my mistake!

    --Chris
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