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Thread: Trigonometric equation

  1. #1
    Newbie Elvis's Avatar
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    Trigonometric equation

    Solve $\displaystyle \sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0$ for $\displaystyle -2\pi \leq x \leq 2\pi$.
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  2. #2
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    Grandad's Avatar
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    Hello Elvis
    Quote Originally Posted by Elvis View Post
    Solve $\displaystyle \sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0$ for $\displaystyle -2\pi \leq x \leq 2\pi$.
    Using $\displaystyle \sin 2x = 2 \sin x \cos x$, write the equation as:
    $\displaystyle \sqrt3\cos 2x -\tfrac12\sin2x = 1$
    Then express the LHS in the form $\displaystyle r\cos(2x+\alpha)$, where:
    $\displaystyle \left\{\begin{array}{l}
    r\cos\alpha = \sqrt3\\
    r\sin\alpha = \tfrac12\\
    \end{array}\right.$
    Can you continue?

    Grandad
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  3. #3
    Newbie Elvis's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello ElvisUsing $\displaystyle \sin 2x = 2 \sin x \cos x$, write the equation as:
    $\displaystyle \sqrt3\cos 2x -\tfrac12\sin2x = 1$
    Then express the LHS in the form $\displaystyle r\cos(2x+\alpha)$, where:
    $\displaystyle \left\{\begin{array}{l}
    r\cos\alpha = \sqrt3\\
    r\sin\alpha = \tfrac12\\
    \end{array}\right.$
    Can you continue?

    Grandad
    Hi, Grandad. My gratitudes.

    $\displaystyle \left\{\begin{array}{l}
    r\cos\alpha = \sqrt3\\
    r\sin\alpha = \tfrac12\\
    \end{array}\right.$

    We have $\displaystyle r\cos\alpha = \sqrt3 \Rightarrow \cos\alpha = \dfrac{\sqrt{3}}{r} $ And $\displaystyle r\sin\alpha = \dfrac{1}{2} \Rightarrow \sin\alpha = \dfrac{0.5}{r}$. Then $\displaystyle r = \sqrt{\left(\sqrt{3}\right)^2+\left(\dfrac{1}{2}\r ight)^2} = \dfrac{\sqrt{13}}{2}$ and $\displaystyle \alpha = \arctan{\left(\dfrac{0.5}{\sqrt{3}}\right)} = 16.10^{\circ}$. Hence $\displaystyle \sqrt3\cos 2x -\tfrac12\sin2x = \dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht)$
    But $\displaystyle \sqrt3\cos 2x -\tfrac12\sin2x = 1$ So $\displaystyle \dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht) = 1 \Leftrightarrow \cos\left(2x+16.10^{\circ}\right) = \dfrac{2}{\sqrt{13}} $ $\displaystyle \Rightarrow 2x+16.10^{\circ} = \arccos{\left(\dfrac{2}{\sqrt{13}}\right)} = 56.30^{\circ} $ or $\displaystyle 326.3^{\circ} $

    $\displaystyle \Rightarrow 2x = 56.30^{\circ}-16.10^{\circ} = 40.2^{\circ}$, or $\displaystyle 326.3^{\circ}-16.10^{\circ} = 310.2^{\circ}$ $\displaystyle \Rightarrow x = \frac{40.2^{\circ}}{2}$, or $\displaystyle \dfrac{310.2^{\circ}}{2}$ $\displaystyle \Rightarrow x = 20.1^{\circ}$ or $\displaystyle 155.1^{\circ} $.

    Hence the solutions of the equation $\displaystyle \sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0
    $ in the interval $\displaystyle [-2\pi, 2\pi] $ are $\displaystyle x = \pm{20.1^{\circ}}$ and $\displaystyle x = \pm{155.1^{\circ}} $. Hope it's right.
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  4. #4
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    Hello Elvis

    Your working is correct down to here:
    Quote Originally Posted by Elvis View Post
    Hi, Grandad. My gratitudes.

    $\displaystyle \left\{\begin{array}{l}
    r\cos\alpha = \sqrt3\\
    r\sin\alpha = \tfrac12\\
    \end{array}\right.$

    We have $\displaystyle r\cos\alpha = \sqrt3 \Rightarrow \cos\alpha = \dfrac{\sqrt{3}}{r} $ And $\displaystyle r\sin\alpha = \dfrac{1}{2} \Rightarrow \sin\alpha = \dfrac{0.5}{r}$. Then $\displaystyle r = \sqrt{\left(\sqrt{3}\right)^2+\left(\dfrac{1}{2}\r ight)^2} = \dfrac{\sqrt{13}}{2}$ and $\displaystyle \alpha = \arctan{\left(\dfrac{0.5}{\sqrt{3}}\right)} = 16.10^{\circ}$. Hence $\displaystyle \sqrt3\cos 2x -\tfrac12\sin2x = \dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht)$
    But $\displaystyle \sqrt3\cos 2x -\tfrac12\sin2x = 1$ So $\displaystyle \dfrac{\sqrt{13}}{2}\cos\left(2x+16.10^{\circ}\rig ht) = 1 \Leftrightarrow \cos\left(2x+16.10^{\circ}\right) = \dfrac{2}{\sqrt{13}} $ $\displaystyle \Rightarrow 2x+16.10^{\circ} = \arccos{\left(\dfrac{2}{\sqrt{13}}\right)} $
    But then you need to say:

    $\displaystyle 2x+16.10=360n\pm 56.30$

    $\displaystyle \Rightarrow x +8.05 = 180n\pm28.15$

    $\displaystyle \Rightarrow x = 180n\pm28.15-8.05$

    Then taking $\displaystyle n = 0$ with the $\displaystyle +$ sign:
    $\displaystyle x = 20.1^o$
    $\displaystyle n = 0$ with the $\displaystyle -$ sign:
    $\displaystyle x =-36.2^o$
    ... etc.

    I think the answers are:
    $\displaystyle x=-339.3^o, -216.2^o, -159.9^o, -36.2^o, 20.1^o, 143.8^o, 200.1^o,323.8^o$

    Grandad
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