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    (16+9sin2x)/(5-3cosx)

    This was the equation we were given, told to put it into form p+qcosx which I got 5+3cosx.
    Then it said to give the smallest angle in radeons, which isn't possible beacuse cos doesn't go to -5/3. Was this an impossible question or am I missing something?
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    (Original post by James73)
    (16+9sin2x)/(5-3cosx)

    This was the equation we were given, told to put it into form p+qcosx which I got 5+3cosx.
    Then it said to give the smallest angle in radeons, which isn't possible beacuse cos doesn't go to -5/3. Was this an impossible question or am I missing something?
    The minimum value for cos(x) is -1, because any value lower than that is impossible (and it asked for the lowest value).

    So what you're trying to do is get cos(x) to equal -1. What you did is made the equation equal to zero.

    So, if we say a = 5 + 3cos(x), we know (a-5)/3 must equal -1. You should have got a as 2. So 2 = 5 + 3cos(x) leading to cos(x) = -1 and therefore x = pi.
 
 
 

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