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Math Help - Trigonometric equation

  1. #1
    Newbie Elvis's Avatar
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    Trigonometric equation

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

    Grandad
    Hi, Grandad. My gratitudes.

    \left\{\begin{array}{l}<br />
r\cos\alpha = \sqrt3\\<br />
r\sin\alpha = \tfrac12\\ <br />
\end{array}\right.

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

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

    Hence the solutions of the equation \sqrt{3}\cos{2x}-\sin{x}\cos{x}-1 = 0<br />
in the interval [-2\pi, 2\pi] are  x = \pm{20.1^{\circ}} and 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.

    \left\{\begin{array}{l}<br />
r\cos\alpha = \sqrt3\\<br />
r\sin\alpha = \tfrac12\\ <br />
\end{array}\right.

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

    2x+16.10=360n\pm 56.30

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

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

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

    I think the answers are:
    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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