# P2 exercise

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Thread starter 15 years ago
#1
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
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15 years ago
#2
(Original post by toxi)
its from the heinemann book, review exercise 1, q 4 part a

solve for 0<=x<=360 the equation

2cos(x+50)=sin(x+40)
This is easy to solve by P3 methods, you dont do compound angle forumla in P2 do you?
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15 years ago
#3
(Original post by 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.
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15 years ago
#4
(Original post by Rustyk1)
This is easy to solve by P3 methods, you dont do compound angle forumla in P2 do you?

how do u do it??
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15 years ago
#5
forget it... i got it......
did it by compound angle formula...lol.....(yes, they're done in p2.)
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15 years ago
#6
sorry, im still a bit baffled as to how that was done!
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15 years ago
#7
(Original post by 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?
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15 years ago
#8
(Original post by 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
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15 years ago
#9
can anyone do p2 examination style paper pg 189 heinemann book question 1?
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15 years ago
#10
(Original post by 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
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15 years ago
#11
(Original post by 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.
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15 years ago
#12
(Original post by 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
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15 years ago
#13
(Original post by 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).
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15 years ago
#14
(Original post by 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.
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15 years ago
#15
(Original post by Ralfskini)
yes, I get the pairs of solutions (3,1) and (1/3,9).
impossible
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15 years ago
#16
(Original post by 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.
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15 years ago
#17
(Original post by Ralfskini)
Nope, they definitely work.
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15 years ago
#18
(Original post by Ralfskini)
Nope, they definitely work.
how?
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15 years ago
#19
(Original post by TheWolf)

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).
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15 years ago
#20
from my book:

(1) = lg x + lg y = 1000
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