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Math Help - trigonometric equation

  1. #1
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    Exclamation trigonometric equation

    How do I start in solving 6sinx = 5 - 8cosx????
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    How do I start in solving 6sinx = 5 - 8cosx????
    One way is you use sin(x) = sqrt[1 -cos^2(x)].
    Another way is you use cos(x) = sqrt[1 -sin^2(x)].

    Or, just square both sides of the posted equation.

    The idea is to arrive to an equation that involves one trig function only.
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  3. #3
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    One way is you use sin(x) = sqrt[1 -cos^2(x)].
    Another way is you use cos(x) = sqrt[1 -sin^2(x)].

    Or, just square both sides of the posted equation.
    I've tried all three but i cannot get the answer out of them (the answers are meant to be 97 degrees and 337 degrees)?
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  4. #4
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    oh no w8 solved it

    squared it wrong
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