Maths Methods 3/4 Help Thread 2011

[brightsky]

cos(2x) = sqrt(3)/2
2x = pi/6, 2pi - pi/6 [These are the first two solutions]
x = pi/12, 11pi/12

[kefoo]

cos(2x) = sqrt(3)/2
2x = pi/6, 2pi - pi/6 [These are the first two solutions]
x = pi/12, 11pi/12

oh so you divide the 2 after u find the first 2 solutions i get it thanks

[thushan]

Wow brightsky if you don't get 50 in everything I will eat myself! You're brilliant - and you're only in year 10!

[onur369]

Ive been saying the same thing, Derrick Ha Jnr

[luffy]

Wow brightsky if you don't get 50 in everything I will eat myself! You're brilliant - and you're only in year 10!

And that includes both Latin and English!!! xD. 3 of his subjects will scale past 50, when he gets his inevitable 50 that is... and then that will supplement for any English study score, provided he doesn't get a 50 in that as well. LOL

Good Luck!

[kefoo]

Hello
i got a question here..

Find the general solution of and find all the solutions for x in the interval (-2pi, 2pi)
2cos(2x + pi/4) = sqrt2

so i find the general eqn which turns out to be x = pi(n)
but i dunno how to find the solutions for x

[brightsky]

2cos(2x + pi/4) = sqrt(2)
cos(2x + pi/4) = sqrt(2)/2
Since x E (-2pi, 2pi)
2x E (-4pi, 4pi)
2x + pi/4 E (-4pi + pi/4, 4pi + pi/4) = (-15pi/4, 17pi/4)
So based on this restriction we get:
x = -2pi - pi/4, -2pi + pi/4, -pi/4, pi/4, 2pi - pi/4, 2pi + pi/4, 4pi - pi/4
Simplify as applicable.

[xZero]

Sub n with integers, n =-1, x=-pi, n =0, x =0, n =1, x =pi

Hence x = -pi,0,pi

[kefoo]

Sub n with integers, n =-1, x=-pi, n =0, x =0, n =1, x =pi

Hence x = -pi,0,pi

yea i tried that but i dont get all the answers
ans are: -5pi/4 , -pi , pi/4, 3pi/4, pi, 7pi/4

[xZero]

Whoops didn't see the 2x there yeh brightsky is  right

[kefoo]

q;
1. If 4logb(|x|) = logb(16) + 8, find x
2. The expression loge(4e^3x) is equal to
3. The expression 3^(log3(x-4)) is equal to

[pi]

Q1








Hope thats right  


Q2






Q3





EDIT: forgot restrictions (pointed out below by m@tty)

[m@tty]

^
Q3. x>4

Other than that it looks right.

[pi]

Q2. x>0

Why is this the restriction? If x=0, the log will still be positive. I'm not sure...

(not contradicting, just confused )

[m@tty]

Q2. x>0

Why is this the restriction? If x=0, the log will still be positive. I'm not sure...

(not contradicting, just confused )

Haha, you can tell me I'm wrong

I read it as being multiplied by x. Clearly this is not the case, and there is no restriction.