Trigonometry

cot^2(3x)sin(2x)-sin(2x)=0

2sin^2(5x)+sin(5x)-1=0

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  1. For the first, take out a common factor of sin(2x) to get
    sin(2x)[cot^2(3x) - 1] = 0
    now you have a difference of squares for
    sin(2x)(cot(3x - 1)(cot(3x) + 1) = 0
    Can you take it from there?


    I will do the second one:
    Your equation is a quadratic which factors.
    (2sin(5x) - 1)(sin(5x) + 1) = 0
    sin(5x) = 1/2 or sin(5x) = -1

    5x = 30º or 150º or 5x = 270º
    x = 6º , 30º , 54º

    I don't know what your domain was, but the period of sin(5x) is 360/5º or 72º
    so by adding or subtracting multiples of 72 to each of my answers will yield as many answers as you want.

    general solutions in radians :
    pi/30 + k(2pi/5)
    pi/6 + k(2pi/5)
    3pi/10 + k(2pi/5) , where k is an integer.

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    posted by Reiny
  2. Thank you! Would you mind explaining to me the k(2pi/5) thing? How do you know when to use that and how do you know when to use just k(pi/5) without the 2? Is the period always going to be the number in front of x (like in this case it was 5 because of 5x)?

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