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    Given that 3 cos 2β + 19 cos β + 13 = 0, where 90◦ < β < 180◦ , find the exact value of secB.

    Factorising and solving gives me :
    CosB = -2/3 and CosB=-5/2
    , which is then converted to 1/cosB=-2/3 hence SecB= -3/2.

    Why is the solutions CosB=-5/2 not included as the solution?
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    If you draw a cos graph you will see that the range for cosx is from 1 to -1 and hence you cannot have the -2.5 value.
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    (Original post by Chelsea12345)
    Given that 3 cos 2β + 19 cos β + 13 = 0, where 90◦ < β < 180◦ , find the exact value of secB.

    Factorising and solving gives me :
    CosB = -2/3 and CosB=-5/2
    , which is then converted to 1/cosB=-2/3 hence SecB= -3/2.

    Why is the solutions CosB=-5/2 not included as the solution?
    -5/2 is less than -1 so cos(β) = -5/2 is an invalid solution.
 
 
 
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