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Thread: 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 $\displaystyle \theta$ in the range $\displaystyle 0^{\circ} \leq \theta \leq 360^{\circ}$ satisfying $\displaystyle 12\ sin^2\ \theta - 5\ cos\ \theta = 9$

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

    Can someone please go through step by step or as much as they can to try and help me?
    Last edited by db5vry; Jan 30th 2009 at 07:35 AM.
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  2. #2
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    Hi

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

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

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

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

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


    And we have:

    . . $\displaystyle \begin{array}{ccccc}3\cos\theta -1\:=\:0 & \Rightarrow & \cos\theta \:=\:\tfrac{1}{3} &\Rightarrow&\theta \:\approx\:70.5^o,\:289.5^o \\ \\

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

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


    . . $\displaystyle \begin{array}{cc}\text{Then:} & 3x + 15^o \:=\:\left\{30^o,\:150^o,\:390^o,\:510^o\right\} \\ \\

    \text{And:} & 3x \:=\:\left\{15^o,\:135^o,\:375^o,\:495^o\right\} \\ \\

    \text{Hence:} & x \:=\:\left\{5^o,\:45^o,\:125^o,\:165^o\right\}\end {array}$

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