



 
Hi Sam, There are different ways to approach trig problems, I often think in terms of their graphs. I see the graph of y = sin(t) for t in [0, 2π], t in radians. You have sin(t) = √3/2 which is approximately 0.8399 so I get two values for t, t_{1} and t_{2}. I also know from the 12√3 triangle that sin(π/3) = √3/2 and hence the distance labeled a in the diagram is π/3. Hence t_{1} = π + π/3 = 4π/3 and t_{2} = 2π  π/3 = 5π/3. Since the sine is periodic with period 2π the solution to sin(t) = √3/2 is t_{1} plus or minus multiples of 2π and t_{2} plus or minus multiples of 2π. In your problem t = 4x so what solutions do you get for sin(4x) = √3/2? Which ones are in [0, 2π]? Harley  


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