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March 25, 2017

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Approximate the equation's solutions in the interval (0,2pi).

sin2x sinx = cosx

I know that I should work on the left side first. I know how to solve for the intervals but I am just not sure how to start this off.

  • Math - ,

    sin2x sinx = cosx
    2sinxcosxsinx - cosx = 0
    cosx(2sin^2x - 1) = 0
    cosx = 0 or sin^2x = 1/2

    if cosx = 0, x = pi/2 or 3pi/2

    if sin^2x = 1/2
    sinx = ± 1/√2
    (x could be in all 4 quadrants)
    x = pi/4, pi-pi/4, pi+pi/4, and 2pi-pi/4
    = ....

    (I don't understand why they want you to approximate, we have exact answers)

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