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Math Help - Solve for angle

  1. #1
    Junior Member
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    Solve for angle

    Greetings.

    I request assistance with solving the following:

    2/cos^2(t) - tan^2(t) = 5


    Would greatly appreciate pointers for solving similar equations.

    Best regards,
    wirefree
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by wirefree View Post
    Greetings.

    I request assistance with solving the following:

    2/cos^2(t) - tan^2(t) = 5


    Would greatly appreciate pointers for solving similar equations.

    Best regards,
    wirefree
    \frac{2}{\cos ^2 x} - \tan ^2x= 5

    \frac{2}{\cos ^2x} - \frac{\sin ^2x}{\cos ^2x}=5

    \frac{2-\sin ^2x - 5\cos ^2x}{\cos ^2x} = 0

    \frac{2-(1-\cos^2 x) - 5\cos ^2 x }{\cos ^2x}=0

    2-1+\cos ^2x - 5\cos ^2x = 0

    4\cos ^2x = 1

    \cos ^2x = \frac{1}{4}

    \cos x = \pm \frac{1}{2}

    take the angle \frac{\pi}{3} in all quarters
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  3. #3
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    Appreciate the prompt response, amer.

    As a follow-up, could you assist me ascertain how to classify the following type of equations:

     <br />
\cos(x) = \sqrt{2 - \sin^2(x)} <br />
=>\\<br />
\cos^2(x) = 2 - \sin^2(x) <br />
=>\\<br />
\sin^2(x) + \cos^2(x) = 2 <br />
=>\\<br />
1 = 2<br />

    What can be stated about such equations?

    Look forward to a prompt response.

    Best regards,
    wirefree
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  4. #4
    MHF Contributor Amer's Avatar
    Joined
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    Jordan
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    Quote Originally Posted by wirefree View Post
    Appreciate the prompt response, amer.

    As a follow-up, could you assist me ascertain how to classify the following type of equations:

     <br />
\cos(x) = \sqrt{2 - \sin^2(x)} <br />
=>\\<br />
\cos^2(x) = 2 - \sin^2(x) <br />
=>\\<br />
\sin^2(x) + \cos^2(x) = 2 <br />
=>\\<br />
1 = 2<br />

    What can be stated about such equations?

    Look forward to a prompt response.

    Best regards,
    wirefree
    ok you can said that this equation can't be correct since sin and cos functions have the values between -1 and 1 .

    cos = \sqrt{2-sin^2 x}

    -1 \leq \sqrt{2-sin^2 x} \leq 1

    0 \leq 2-\sin ^2 x \leq 1

    -2 \leq -\sin ^2 x \leq -1 !!! contradiction

    1 \leq \sin ^2 x \leq 2 !!! contradiction

    since

    0 \leq \sin ^2 x \leq 1
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  5. #5
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    Get the cos(x) on the other side of the equation with a minus so you get the big square root minus Cos(x) = 0

    So you try to find an x for which this f(x) is worth zero. Just by looking at the function I can see its always positive...if you don't believe me try to find a x value for which F(x) is negative, you won't find any.

    Just say that there is no x for which this relation is verified or you can calculate the derivative of the function and find the minimum ( should be around 0.5 eye wise). I think the period is huge as well like modulo a billion pie LOL
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