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Math Help - Help with evaluating trigonometric equations...

  1. #1
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    Help with evaluating trigonometric equations...

    I am self-studying A-levels pure maths and I need a little bit of help with understanding these questions.

    Find, to 1 decimnal place, all values of z in the interval -180 <= z <= 180 satisfying:

    (a) coz z deg (1 + sin z deg) = 0

    In this questions, it's easy to guess that it must be at -90 or 90 but I don't understand how to solve the equation.

    (b) (1 - tan z deg) sin z deg = 0

    I am clueless on this one.

    Ideas please? Perhaps could you help me find a place that better explains this?
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  2. #2
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    Lexington, MA (USA)
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    Hello, struck!

    You forgot the most basic rule for solving equations . . .

    . . If a\cdot b \:=\:0, then a = 0 or b = 0.


    If a product equals zero, then one of the factors must be zero.


    Find, to 1 decimal place, all values of z in [\text{-}180^o,\:180^o] satisfying:

    (a)\;\;\cos z(1 + \sin z) \:=\: 0
    So we have:

    . . \cos z \:=\:0 \quad\Rightarrow\quad\boxed{ z \;=\;\text{-}90^o,\:90^o}

    . . 1 + \sin z \:=\:0 \quad\Rightarrow\quad \sin z \:=\:-1 \quad\Rightarrow\quad\boxed{ z \:=\:\text{-}90^o}




    (b)\;\;(1 - \tan z)\sin z \:= \:0
    So we have:

    . . 1-\tan z \:=\:0 \quad\Rightarrow\quad \tan z \:=\:1 \quad\Rightarrow\quad\boxed{ z \;=\;45^o,\:\text{-}135^o}

    . . \sin z \:=\:0 \quad\Rightarrow\quad\boxed{ z \;=\;\text{-}180^o,\: 0^o,\:180^o}

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