The answer to this is (7/6, - root 3)

I keep getting (5/6, - root 3)

Anyone know where I am going wrong?

## 2002 Paper 1 Q.8b

### dfx

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)