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Double Angle Trig

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Sin2x-cosx=1
    Solve for all x values between [0,2pi)

    2. Relevant equations
    Sin2x=2sinxcosx

    3. The attempt at a solution

    2sinxcosx-cosx=1
    cosx(2sinx-1)=1

    I don't know what to do after this. It doesn't equal 0 so I can't set each factor equal to 0
     
  2. jcsd
  3. Mar 5, 2015 #2

    Mark44

    Staff: Mentor

    Are you sure that the problem isn't sin2(x) - cos(x) = 1? If the problem is exactly as you have stated, I don't know where to go, either.
     
  4. Mar 5, 2015 #3
    The problem is indeed sin(2x). It's really giving me a headache.
     
  5. Mar 5, 2015 #4

    Mark44

    Staff: Mentor

    There appear to be solutions at x = ##\pm\pi## and many other points (from wolframalpha) but it's not clear to me how to get them.
     
  6. Mar 5, 2015 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Remember, cos θ = sin (θ + π/2).

    I think you can use this identity and get an expression for the LHS involving only the sine of the angle θ.

    After that, we can talk some more. :smile:
     
  7. Mar 5, 2015 #6

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Alternatively, you can plot ##\sin(2x)## and ##1+\cos x## on the same graph for ##0\le x\le 2\pi##. They apparently cross at ##\pi## and ##\frac{3\pi} 2##, which are both easily verified.
     
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