I’ve been knocking out the trig problems in this section with minimal difficulty so far, but I’ve run straight into a brick wall on this “Algebraic” part. I’m asked to find sin(x)=0 between [0,2π). If I graphed the unit circle this would be a trivial exercise to show sin(θ)=0 when θ=0 or π.

Where I have trouble is- I’m very explicitly being told here that the solution is ALGEBRAIC, and I’m struggling to figure out a way to rearrange sin(x)=0 to come up with the known answer. Further, unit circles are not in this chapter, they wouldn’t likely ask me to exercise a skill taught in another chapter. What am I missing?

It’s not just 31, either. Looking ahead at eg 37, I can easily show sin(-x) = -sin(x) on a unit circle. I could maybe fuck around with inverse trig ratios but those are in section 3- this is only section 1.

Help me out here, drop a hint, share a link: how do I solve sin(x)=0 on [0,2π), but algebraically? I suspect it’s something glaringly obvious and/or very very simple I’ve overlooked.

  • DarkNightoftheSoulOPEnglish
    1·
    10 months ago

    Okay, like I get that, but I don’t think that’s the procedure intended by the author. For one, it will only give me one of the two correct solutions in this range domain. For another, I’m not supposed to know about inverse functions yet, that’s two sections ahead.

    • jdnewmil@lemmy.ca
      2·
      10 months ago

      Nope. When you use sqrt to solve y=x^2 , you are expected to recognize that sqrt is only the inverse of the right branch of the parabola. Likewise, arcsin is only the inverse of the sine limited to +/-pi/2, so you have to use it intelligently as a tool, not blindly as a black box.

      • myslsl@lemmy.world
        2·
        10 months ago

        The point they made was correct. Arcsine by itself only gives one of the two solutions to sinx=0. It seems like they already realize that they need to use arcsine carefully if they use it at all.