Quick FP3 Complex Number Question Watch

simoncb
Badges: 1
#1
Report Thread starter 8 years ago
#1
Could someone please help me with a quick question... I have done 95% of the question but dont get 1 small bit:

We show that cos^5(x) = 1/16(cos5x + 5cos3x + 10cosx) in part 1.
Then they say Hence solve the equation cos5x + 5cos3x + 9cosx = 0 for x between 0 and Pi

So I can see they let 16cos^5(x) = cosx to get from the first equation to the 2nd... therefore cosx = +/- 0.5, and x = Pi/3 and 2Pi/3

But in the mark scheme, there is one more value of x, which they get by saying cosx = 0, so x = Pi/2

How do they get that last value of x? (i.e where do they get cosx = 0 from?)
0
reply
DFranklin
Badges: 18
Rep:
?
#2
Report 8 years ago
#2
16x^5 - x = x(16x^4 - 1) = x(4x^2+1)(2x-1)(2x+1).

Note that x is still a factor, so x = 0 is a solution.
0
reply
nuodai
Badges: 14
#3
Report 8 years ago
#3
(Original post by simoncb)
How do they get that last value of x? (i.e where do they get cosx = 0 from?)
I'm guessing you made the mistake of dividing through by \cos x; in doing so you lost the root \cos x = 0. For example, if you had 3x^2 - x = 0 you'd factorise to get x(3x-1) = 0, you wouldn't just divide through to get 3x-1=0 because then you lose a root.

So instead of dividing through you need to gather terms on one side and factorise. Then the \cos x = 0 root becomes a lot more clear.
reply
IDGAF
Badges: 0
Rep:
?
#4
Report 8 years ago
#4
(Original post by simoncb)
Could someone please help me with a quick question... I have done 95% of the question but dont get 1 small bit:

We show that cos^5(x) = 1/16(cos5x + 5cos3x + 10cosx) in part 1.
Then they say Hence solve the equation cos5x + 5cos3x + 9cosx = 0 for x between 0 and Pi

So I can see they let 16cos^5(x) = cosx to get from the first equation to the 2nd... therefore cosx = +/- 0.5, and x = Pi/3 and 2Pi/3

But in the mark scheme, there is one more value of x, which they get by saying cosx = 0, so x = Pi/2

How do they get that last value of x? (i.e where do they get cosx = 0 from?)
When solving this sort of equation you have to make sure you don't divide through by cosx, as you then lose the solution cosx = 0

When you have 16cos^5x = cosx, this should become 16cos^5x - cosx = 0

Then factorise. That should help you.
0
reply
simoncb
Badges: 1
#5
Report Thread starter 8 years ago
#5
Ah of course... thanks the 3 of you! Silly error really
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the Arts London
    MA Performance Design and Practice Open Day Postgraduate
    Thu, 24 Jan '19
  • Brunel University London
    Undergraduate Experience Days Undergraduate
    Sat, 26 Jan '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Sat, 26 Jan '19

Are you chained to your phone?

Yes (57)
19.39%
Yes, but I'm trying to cut back (119)
40.48%
Nope, not that interesting (118)
40.14%

Watched Threads

View All
Latest
My Feed

Oops, nobody has posted
in the last few hours.

Why not re-start the conversation?

Start new discussion

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise