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.

  • jdnewmil
    2·
    9 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·
      9 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.