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You can type it out like I do, and that takes a bit of getting used to. Alternatively you could go here: http://www.codecogs.com/latex/eqneditor.php and enter whatever you want then paste it back here and putting the tags [ tex ] and [ /tex ] at the beginning and end. Or a third method is type...

Oops, too fast smiley :P

tan(2\theta) = \pm \sqrt{3} \\ 2\theta = 60^\circ, (180^\circ-60^\circ), 240^\circ, 300^\circ, -(180^\circ-60^\circ), -60^\circ, -300^\circ, -240^\circ \\ \therefore \theta = 30, 60, 120, 150, -60, -30, -150, -120 \\ (\pm 30, \pm 60, \pm 120, \pm 150) I think that's correct I'm not too sure.

The domain is 0^{\circ} \leq \theta \leq 360^{\circ} So its not values of one theta between 0 and 180 degrees, but one theta between 0 and 360 as the question states. What you should be getting is 2\theta = 60, 300, 420, 660 \\ \therefore \theta = 30, 150, 210, 330

So how did you get -315' when you add +360 for the second revolution? Sorry for all the questions :\

Yes I already knew that, but my main problem here is, since you do a second revolution don't you add say, 45 + 360 = 405 In my first revolution I get 2x = 45 and -135. In my second revolution I get (45+360=405) and (-135+360=225)

It's -67'30', I just worked it out myself and got the right answer, but now I'm trying to explain to myself why it works :\

Ah nope sorry, its tan2x = 1 not 2tanx = 1. The answers are 22.5, 112.5, -67.5 and -157.5 degrees.

It was from a maths in focus text book.

Hey guys, Could you explain how to work this out? $Solve for -180^{\circ} \leq \Theta \leq 180^{\circ} tan(2{\Theta}) = 1$ Thank you :)

The gradient of the normal is -1/x not -x.

Ehh, task #3 being what exactly?

No worries ;), you too.

Whoops forgot to answer back sorry xD You're on the right track, it helps to graph out the three points so its easier. Find the distance between Q and U, this will give you your height. Find the distance between U and P, this will give you your base. Then use A = 1/2 bh. (You don't find the...

Equation of tangent: dy/dx = 3x^2 m = 3(1)^2 = 3 y-1 = 3(x-1) 3x-y-2=0 Equation of the normal: m = -1/3 y-1 = -1/3(x-1) 3y - 3 = -x+1 x+3y-4=0 --- Coordinates of P: (where the tangent meets the y-axis) sub x = 0 into the equation of the tangent 3(0)-y-2=0 y=-2 Therefore P(0,-2)...

Sorry there you go.

Well luckily for you, you don't actually need circle geo :P. Here's my working out: (Sorry for the bad quality, scanner wasn't working so I just took a photo of it)

No. There are actually two lines that pass through the point and are a tangent to the curve at points (6,36) and (2,-12). If you figure out the equation of the two lines, you'll see that one has a positive gradient and the other has a negative gradient.