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Thread: Solving Trigonometric Equations.

  1. #1
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    Solving Trigonometric Equations.

    Solve the trigonometric equation. For each find the exact solutions in [0,2pi]

    A. Sin(3x)=-1/2
    B.COS (2x)=COS (x)
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  2. #2
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    Re: Solving Trigonometric Equations.

    Hey Cristina999.

    What have you tried?
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  3. #3
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    Re: Solving Trigonometric Equations.

    "Sine", if I recall correctly, is "opposite side over hypotenuse" so I might start by drawing a right triangle with leg opposite the angle equal to half the hypotenuse. Then I might realize that If I put two of those together, with the third sides as common side, I have a triangle with all three sides the same length- two sides are the two hypotenuses and the third side is made with the two 1/2 sides. What can you say about the angles in a equilateral triangle?

    For the other problem, what do you know about when two angles have the same cosine? Think about the graph of y= cos(x).
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  4. #4
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    Re: Solving Trigonometric Equations.

    from your intimate knowlege of the unit circle, you should already know that sine has a value of $-\dfrac{1}{2}$ at the angles $\dfrac{7\pi}{6}$ in quad III and $\dfrac{11\pi}{6}$ in quad IV.

    $0 \le x < 2\pi \implies 0 \le 3x < 6\pi$

    $3x = \bigg[\dfrac{7\pi}{6} \, , \, \dfrac{11\pi}{6} \, , \, \dfrac{19\pi}{6} \, , \, \dfrac{23\pi}{6} \, , \, \dfrac{31\pi}{6}\, , \, \dfrac{35\pi}{6} \bigg]$

    finish solving for $x$.


    For the second problem, note the double angle identity $\cos(2x)=2\cos^2{x}-1$ ...

    $2\cos^2{x}-1=\cos{x}$

    $2\cos^2{x}-\cos{x}-1=0$

    $(2\cos{x}+1)(\cos{x}-1)=0$

    use the zero product property ...
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