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Math Help - solution in the interval [0,360] cos2theta=square root (3)/2

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    solution in the interval [0,360] cos2theta=square root (3)/2

    need help solving
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    \displaystyle \cos 2\theta =\frac{\sqrt{3}}{2}

    Make \displaystyle  2\theta =x therefore

    \displaystyle \cos x =\frac{\sqrt{3}}{2}

    Using the 30-60-90 special triangle

    \displaystyle  x =\frac{\pi}{6}\implies 2\theta = \frac{\pi}{6} \implies \theta = \frac{\pi}{12}

    Now what other quadrants is \cos positive in? and how many solutions are we looking for?
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    Quote Originally Posted by idunit14 View Post
    need help solving
    If you want to use a method that does not require as much analysis...

    \displaystyle \cos{2\theta} = \frac{\sqrt{3}}{2}

    \displaystyle 2\theta = \{30^{\circ}, 360^{\circ} - 30^{\circ}\} + 360^{\circ}n where \displaystyle n is an integer

    \displaystyle 2\theta = \{30^{\circ}, 330^{\circ}\} + 360^{\circ}n

    \displaystyle \theta = \{15^{\circ}, 165^{\circ}\} + 180^{\circ}n.


    So the solutions in the domain \displaystyle [0, 360] are

    \displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}.
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    Quote Originally Posted by idunit14 View Post
    need help solving
    If you want to use a method that does not require as much analysis...

    \displaystyle \cos{2\theta} = \frac{\sqrt{3}}{2}

    \displaystyle 2\theta = \{30^{\circ}, 360^{\circ} - 30^{\circ}\} + 360^{\circ}n where \displaystyle n is an integer

    \displaystyle 2\theta = \{30^{\circ}, 330^{\circ}\} + 360^{\circ}n

    \displaystyle \theta = \{15^{\circ}, 165^{\circ}\} + 180^{\circ}n.


    So the solutions in the domain \displaystyle [0, 360] are

    \displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}.
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