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Math Help - How to Solve Trigonmetric Equation

  1. #1
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    How to Solve Trigonmetric Equation

    Hi, I have two equations each which two variables:

    x*cos (30) - 275*cos(a) = 0

    x*(sin 30) + 275*sin(a) - 300 = 0

    I tried solving for the unknowns but eventually I end up with this equation and get stuck.

    158.77*cos(a) + 275*sin(a) - 300 = 0

    Can someone help me with this? I'm not very good at trigonometric identities...
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by zerobladex View Post
    Hi, I have two equations each which two variables:

    x*cos (30) - 275*cos(a) = 0

    x*(sin 30) + 275*sin(a) - 300 = 0

    I tried solving for the unknowns but eventually I end up with this equation and get stuck.

    158.77*cos(a) + 275*sin(a) - 300 = 0

    Can someone help me with this? I'm not very good at trigonometric identities...
    Note: \cos30^\circ=\sqrt{3}{2} and \sin30^\circ=\frac{1}{2}

    Solving for x in the first equation gives x=\frac{275\cos{a}}{\sqrt3/2}

    Substituting into the second equation

    \left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}+275\sin{a}=0

    Now, we want everything in terms of one trig function, so we use \sin{a}=\pm\sqrt{1-\cos^2a}, then

    \left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}+275(\pm\sqrt{1-\cos^2a})=0

    \left[\left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}\right]^2=\left(275(\pm\sqrt{1-\cos^2a})\right)^2

    can you finish?
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  3. #3
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    I'm still not sure how to solve it, since in the original equation there was a 300 too.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by zerobladex View Post
    I'm still not sure how to solve it, since in the original equation there was a 300 too.
    Sorry, I mssed that. OK, then:

    \left[\left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}{\color{blue}-300}\right]^2=\left(275(\pm\sqrt{1-\cos^2a})\right)^2

    How bout now?

    And it may help if you let u=\cos{a}, such that

    \left[\left(\frac{275{\color{red}u}}{\sqrt3/2}\right)\frac{1}{2}-300\right]^2=\left(275(\pm\sqrt{1-{\color{red}u}^2})\right)^2
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