sarah99
Hey @sarah99,
Sorry if it's too late but to answer that I'll go through the whole working out. So to summarise the question, it's solving the equation π¦ = 2 sin(π₯)β1 for π₯ where π¦ = 0, over the domain 0β€π₯β€3π.
2sin(π₯)β1=0
sin(π₯)=1/2
π₯=π/6 (which is correct with your reference angle you calculated too)
So from there, you need to first draw out or see what x values equal to π/6 from 0 to 2π first. So obviously, that would be π₯=π/6 in the first quadrant and then π₯=5π/6 in the second quadrant (remember, these are quadrants where sin is positive). The other two quadrants in 0 to 2π don't have values for x since they will be negative.
Now extending this to 3π, you see that you are essentially extending to 180 degrees (or π). So here you get another two solutions that are in the first and second quadrant as well which will be essentially 2π plus the first two solutions we got so that will be π₯=2π+π/6=13π/6 and π₯=2π+5π/6=17π/6.
So yeah all your solutions for π₯ will be π/6, 5π/6, 13π/6 and 17π/6. Since there are two sets of quadrant 1 and 2 from 0 to 3π.
Anyways, hope that helps and let me know if you have any more questions!
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