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Trigonometry - How to solve this equation and find the solutions.

  1. Jan 17, 2012 #1
    Solve: sin^2x = (sinx)(cosx)
    0 < x < 2pi

    Find exact solutions.

    Okay, so I got 0, pi/4, but just wasn't sure how to do this...

    I know you bring the sinxcosx to the other side, making it

    sin^2x - sinxcosx = 0

    ...and then I tried factoring out sinx leaving me with sinx(sinx - cosx). That's where I got lost. So sinx is 0 at 0 and pi on the unit circle, so two solutions are 0 and pi, but what on earth do I do with sinx - cosx? That's when I started thinking about possible other identities that could help me out, but each left me with more craziness on my page!
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  3. Jan 17, 2012 #2


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    [itex]\sin x(\sin x - \cos x) = 0[/itex]

    If [itex]a*b=0[/itex] then either a = 0 OR b = 0 (or both can be zero).

    You've covered the first part (equating [itex]\sin x[/itex] to zero) and found all the solutions).

    For the second part, equate [itex]\sin x - \cos x = 0[/itex].

    The easiest way to go from here is to divide throughout by [itex]\cos x[/itex] then rearrange the equation. But take care that by doing this, you're disallowing solutions that make [itex]\cos x = 0[/itex] (because then you'd be dividing by 0). So check that none of those satisfy the original equation.

    EDIT: In this case, it's not strictly necessary to do this (the check I mentioned), but it's good practice when an equation is solved by cancelling out factors from both sides (rather than factorising the entire expression).
    Last edited: Jan 18, 2012
  4. Jan 17, 2012 #3


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    Another way to solve

      cos(x) - sin(x) = 0

    is to multiply both sides by cos(x) + sin(x). Then recognize the identity for cos(2x). With this method you have to check for extraneous solutions.

    (Next someone will suggest the exact way to solve this.)
  5. Jan 17, 2012 #4


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    Well, OK, if you twist my arm. :biggrin:

    [itex]\sin x - \cos x = 0 \Rightarrow \sqrt{2}\sin(x - \frac{\pi}{4}) = 0 \Rightarrow \sin(x - \frac{\pi}{4}) = 0[/itex]
  6. Jan 17, 2012 #5


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    The equation sinx-cosx =0 is equivalent with sinx=cosx. There are two angles between 0 and 2pi having equal sine and cosine.

    Last edited: Jan 18, 2012
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