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Math Help - trig problem

  1. #1
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    trig problem

    This is on a C2 exam past paper and I haven't studied this but I want to know how to do it:

    It is a two part question:

    a] Find all values of \theta in the range 0^{\circ} \leq \theta \leq 360^{\circ} satisfying 12\ sin^2\ \theta - 5\ cos\ \theta = 9

    b] Find all values of x in the range 0^{\circ} \leq x \leq 180^{\circ} satisfying sin\ (3x+15^{\circ})\ =\ 0.5

    Can someone please go through step by step or as much as they can to try and help me?
    Last edited by db5vry; January 30th 2009 at 08:35 AM.
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  2. #2
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    Hi

    a] Replace \ sin^2 \theta by 1 - \cos^2\theta and you will get a second degree equation in \cos \theta

    b] \sin (3x+15^{\circ}) = 0.5 = \sin 30^{\circ}
    \sin a = \sin b leads to a = b + 360^{\circ}k or a = 180^{\circ} - b + 360^{\circ}k with k real
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  3. #3
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    Hello, db5vry!

    (a) Solve: .  12\sin^2\!\theta - 5\cos\theta \:=\:0,\,\text{ for }0^o \leq \theta \leq 360^o

    We have: . 12(1-\cos^2\!\theta) - 5\cos\theta \:=\:0 \quad\Rightarrow\quad 12\cos^2\!\theta + 5\cos\theta - 3 \:=\:0

    Factor: . (3\cos\theta-1)(4\cos\theta + 3) \:=\:0


    And we have:

    . . \begin{array}{ccccc}3\cos\theta -1\:=\:0 & \Rightarrow & \cos\theta \:=\:\tfrac{1}{3} &\Rightarrow&\theta \:\approx\:70.5^o,\:289.5^o \\ \\<br /> <br />
4\cos\theta + 3 \:=\:0 &\Rightarrow& \cos\theta \:=\:\text{-}\tfrac{3}{4} & \Rightarrow & \theta \:\approx\:138.6^o,\:221.4^o\end{array}




    (b) Solve: . \sin(3x+15^o) \:=\:\tfrac{1}{2},\;\text{ for }0^o \leq x \leq 180^o

    We have: . \sin(3x+15^o) \:=\:\tfrac{1}{2}


    . . \begin{array}{cc}\text{Then:} & 3x + 15^o \:=\:\left\{30^o,\:150^o,\:390^o,\:510^o\right\} \\ \\<br /> <br />
\text{And:} & 3x \:=\:\left\{15^o,\:135^o,\:375^o,\:495^o\right\} \\ \\<br /> <br />
\text{Hence:} & x \:=\:\left\{5^o,\:45^o,\:125^o,\:165^o\right\}\end  {array}

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