The answer to this is (7/6, - root 3)
I keep getting (5/6, - root 3)
Anyone know where I am going wrong?
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2002 Paper 1 Q.8b
Started by gary, May 15 2005 12:28 PM
1 reply to this topic
#2
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Posted 16 May 2005 - 01:57 PM
y = 2Cos(2x)
2Cos(2x) = -rt.3
Cos(2X) = -rt.3 / 2
Cos inv. of rt.3/2 = 30 and cos is negative in the 2nd quad (180 - 30) and 3rd quad (180 + 30)
So 2x = 150, 210, ... you could add on 360 but it'd be out of range anyway.
So x = 75, 105
x = 5pi/12, 7pi/12
But the first point of intersection 5pi/12 occurs at A, so B has to intersect at (7pi/12, -rt.3)
2Cos(2x) = -rt.3
Cos(2X) = -rt.3 / 2
Cos inv. of rt.3/2 = 30 and cos is negative in the 2nd quad (180 - 30) and 3rd quad (180 + 30)
So 2x = 150, 210, ... you could add on 360 but it'd be out of range anyway.
So x = 75, 105
x = 5pi/12, 7pi/12
But the first point of intersection 5pi/12 occurs at A, so B has to intersect at (7pi/12, -rt.3)
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