P2 exercise

toxi

solve for 0<=x<=2π the equation

cos2x = 2sin^2 x

Use cos 2x = 1 - 2sin^2 x:

1 - 2sin^2 x = 2sin^2 x

=> 4sin^2 x = 1

=> sin^2 x = 1/4

=> sin x = +-(1/2)

Go from there.

Reply 3

[_Z_]

Rustyk1

This is easy to solve by P3 methods, you dont do compound angle forumla in P2 do you?

how do u do it??

Reply 4

[_Z_]

forget it... i got it...... :biggrin:

did it by compound angle formula...lol.....(yes, they're done in p2.)

Reply 5

jazzyj

sorry, im still a bit baffled as to how that was done!

Reply 6

toxi

its from the heinemann book, review exercise 1, q 4

solve for 0<=x<=360 the equation

2cos(x+50)=sin(x+40)

solve for 0<=x<=2π the equation

cos2x = 2sin^2 x

no idea how to do the first bit - anyone?

Reply 7

lgs98jonee

2

toxi

its from the heinemann book, review exercise 1, q 4

solve for 0<=x<=360 the equation

2cos(x+50)=sin(x+40)

2cosxcos50-2sinxsin50=sinxcos40+cosxsin40
2cos50-2tanxsin50=tanxcos40+sin40
so 2cos50-sin40=tanx(cos40+2sin50)
so tan x= (2cos50-sin40)/(cos40+2sin50)
so x= tan-1 of all that

Reply 8

can anyone do p2 examination style paper pg 189 heinemann book question 1?

Reply 9

lgs98jonee

2

TheWolf

can anyone do p2 examination style paper pg 189 heinemann book question 1?

NOOOOO i was just about to post as well...i think it is impossible in fact almost certain...if u put in the 'solutions' from the back the equations dont work

Reply 10

Hoofbeat

14

lgs98jonee

NOOOOO i was just about to post as well...i think it is impossible in fact almost certain...if u put in the 'solutions' from the back the equations dont work

I asked about that question a while ago. There's a printing error in the question, and we were trying to work out what it should have said, but we couldn't. So I'm afraid you'll just have to leave that question. Normally, the methods involve getting rid of the logs and sovling simultaneously like normal.

Reply 11

lgs98jonee

2

Hoofbeat

I asked about that question a while ago. There's a printing error in the question, and we were trying to work out what it should have said, but we couldn't. So I'm afraid you'll just have to leave that question. Normally, the methods involve getting rid of the logs and sovling simultaneously like normal.

yeh tx

Reply 12

Ralfskini

TheWolf

can anyone do p2 examination style paper pg 189 heinemann book question 1?

yes, I get the pairs of solutions (3,1) and (1/3,9).

Reply 13

IntegralAnomaly

Ralfskini

yes, I get the pairs of solutions (3,1) and (1/3,9).

substitute them into the first eqn lgx+lgy=1000
welll lgx+lgy=lg(xy) , so lg(xy)=1000.
if x=3 and y=1 then lg(3)=1000? no.
and if x=1/3 and y=9 then again lg3=1000.

Reply 14

Ralfskini

yes, I get the pairs of solutions (3,1) and (1/3,9).

impossible

Reply 15

Ralfskini

IntegralAnomaly

substitute them into the first eqn lgx+lgy=1000
welll lgx+lgy=lg(xy) , so lg(xy)=1000.
if x=3 and y=1 then lg(3)=1000? no.
and if x=1/3 and y=9 then again lg3=1000.

Nope, they definitely work.

Reply 16

Ralfskini

Nope, they definitely work.

do your working please

Reply 17

IntegralAnomaly

Ralfskini

Nope, they definitely work.

how?

Reply 18

Ralfskini

TheWolf

do your working please

lgx+lgy=lg3 (1)
lg(3x+y)=1 (2)

From (1): xy=3 ---> y=3/x

lg(3x+y)=1, so 3x+y=10.

3x+3/x=10
3x^2-10x+3=0
(3x-1)(x-3)=0
x=1/3, x=3
when x=1/3, y=9
when x=3, y=1

(and they worked when I checked them).

Reply 19