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Math Help - Trigonometric Equation Problem

  1. #1
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    Trigonometric Equation Problem

    4sinx-2cos(2x)=1

    Please help, how to solve this equation
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  2. #2
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    Re: Trigonometric Equation Problem

    I would start by using the trig identity cos(2x)= cos^2(x)- sin^2(x)= (1- sin^2(x)- sin^2(x))= 1- 2sin^2(x)

    Using that, 4 sin(x)- 2 cos(2x)= 4 sin(x)- 2+ 2sin^2(x)= 1 so that the equation is the same as 2sin^2(x)+ 4sin(x)- 3= 0. If you let y= sin(x), that is 2y^2+ 4y- 3= 0. Can you solve that quadratic equation for y and then solve for x?
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  3. #3
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    Re: Trigonometric Equation Problem

    Thanks for answering


    I solved quadratic equation 4u^2+4u-3=0 (sinx=u)

    u1=1/2
    u2=-1.5 (dont satisfy)

    so sinx=1/2 but i need to write two separate answers, what x1 and x2 are, but i get this

    x=(-1)^n * Pi/6 + 2Pin

    how to write two answers from that one, which i learned? please help
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  4. #4
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    Re: Trigonometric Equation Problem

    Quote Originally Posted by Telo View Post
    Thanks for answering
    I solved quadratic equation 4u^2+4u-3=0 (sinx=u)
    u1=1/2
    u2=-1.5 (dont satisfy)
    so sinx=1/2 but i need to write two separate answers, what x1 and x2 are, but i get this
    x=(-1)^n * Pi/6 + 2Pin
    Actually the two principal values are \frac{\pi}{6}~\&~\frac{5\pi}{6}
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  5. #5
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    Re: Trigonometric Equation Problem

    can you please explain how did you get those answers?
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  6. #6
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    Re: Trigonometric Equation Problem

    Quote Originally Posted by Telo View Post
    can you please explain how did you get those answers?
    \sin\left(\frac{\pi}{6}\right)=0.5~\&~\sin\left( \frac{5\pi}{6}\right)=0.5
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  7. #7
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    Re: Trigonometric Equation Problem

    Quote Originally Posted by Telo View Post
    can you please explain how did you get those answers?
    are you not familiar with the unit circle?

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