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Solve the equation for 0<x<2π
√(2) sin(x)+√(2) cos(x)>0

  • math -

    divide by √2
    sinx + cosx > 0
    sinx > -cosx
    sinx/cosx > -1
    tanx > -1

    from the graph of the standard tangent curve of y = sinx and y = -cosx

    in the domain 0 < x < 2π we see that sinx is above -cosx for
    0 < x < 3π/4 OR 7π/4 < x < 2π

  • use 2nd solution - math -

    revised solution:


    divide by √2
    sinx + cosx > 0
    sinx > -cosx


    from the graph of the standard curve of y = sinx and y = -cosx

    in the domain 0 < x < 2π we see that sinx is above -cosx for
    0 < x < 3π/4 OR 7π/4 < x < 2π



    I originally went with tanx, as seen in my first attempt,
    obtained by dividing both sides by cosx
    However, since division by cosx would result in worrying about positive and negatives divisors giving me reversals of the inequality sign, I just went with the second version of my solution

  • math -

    or, recognize that what you have is

    2sin(x+pi/4) > 0

    so, x+pi/4 must be in QI or QII
    -pi/4 < x < 3pi/4
    But, we want x>0, so add 2pi to make it positive:
    -pi/4 < x is the same as 7pi/4 < x < 2pi

    answer is as shown above.

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