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    How do you draw the angle?

    E.g. with bounds between 0 and 2pi for a solution of pi in the S and T segments

    (2cosx-1)(cosx+1)=0

    im looking for the one where cosx+1 = 0
    cosx = -1 (-ve so it has to be in S and T segment).

    Please could someone upload a diagram of what the circle should look like, thanks
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    When cos(x) or sin(x) is equal to +1 or -1, I would always recommend drawing a sketch graph of the function to get an idea of where you are (and where the repeat solutions might be). In the case you've quoted, i.e. cos(x) = -1, the result a calculator will respond with is 180 degrees or pi radians. You should be aware that the sin and cos function repeat every 360 degrees (2pi radians), so the immediately neighbouring solutions to that given by the calculator will be 180-360 = -180, or 180+360 = 540. Both of these are outside your target range of 0 to 2pi.

    On the CAST diagram, the angle 180 degrees (pi radians) sits on the line that bisects the S and T quadrants, i.e. it is on the negative horizontal axis. Hope this will do in the absence of a diagram.
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    (Original post by old_engineer)
    When cos(x) or sin(x) is equal to +1 or -1, I would always recommend drawing a sketch graph of the function to get an idea of where you are (and where the repeat solutions might be). In the case you've quoted, i.e. cos(x) = -1, the result a calculator will respond with is 180 degrees or pi radians. You should be aware that the sin and cos function repeat every 360 degrees (2pi radians), so the immediately neighbouring solutions to that given by the calculator will be 180-360 = -180, or 180+360 = 540. Both of these are outside your target range of 0 to 2pi.

    On the CAST diagram, the angle 180 degrees (pi radians) sits on the line that bisects the S and T quadrants, i.e. it is on the negative horizontal axis. Hope this will do in the absence of a diagram.
    Name:  x.png
Views: 69
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    Like the red line ive drawn?

    Why doesnt the green line work as a solution (i know it ends in the C segment) - but it starts from the T segment
 
 
 
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