Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    3
    ReputationRep:
    So confused

    Question 3B asks

    Find, in terms of π, the solutions of the equation sin 5x + sin x = 0, for x in the interval 0 ≤ x < π.

    I didn't know how to do it, but the mark scheme gave this as the identity you need to turn the equation into?

    2 sin 3x cos 2x = 0


    How do you form this equation what identity is it using?

    Links to paper and MS

    https://b3755649dbd1afe3db91a899c3b9...%20Edexcel.pdf


    https://b3755649dbd1afe3db91a899c3b9...%20Edexcel.pdf
    Offline

    22
    ReputationRep:
    (Original post by eyeman567)
    So confused

    Question 3B asks

    Find, in terms of π, the solutions of the equation sin 5x + sin x = 0, for x in the interval 0 ≤ x < π.

    I didn't know how to do it, but the mark scheme gave this as the identity you need to turn the equation into?

    2 sin 3x cos 2x = 0


    How do you form this equation what identity is it using?

    Links to paper and MS

    https://b3755649dbd1afe3db91a899c3b9...%20Edexcel.pdf


    https://b3755649dbd1afe3db91a899c3b9...%20Edexcel.pdf
    You need to realise there is a link between Q3(a) and Q3(b).

    From question 3(a), if you set P=5x and Q=x you get:

    \displaystyle 

\begin{equation*}\sin 5x + \sin x = 2 \sin \frac{5x + x}{2} \cos \frac{5x-x}{2} = 2\sin 3x \cos 2x\end{equation*}.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    You need to realise there is a link between Q3(a) and Q3(b).

    From question 3(a), if you set P=5x and Q=x you get:

    \displaystyle 

\begin{equation*}\sin 5x + \sin x = 2 \sin \frac{5x + x}{2} \cos \frac{5x-x}{2} = 2\sin 3x \cos 2x\end{equation*}.
    Thanks a lot haha, missed that completely.

    could you help me with 4a?


    y = x^5/2 ln x/4

    when Y = 0 find the coordinate, only one mark but unsure what I need to do
    Offline

    22
    ReputationRep:
    (Original post by eyeman567)
    Thanks a lot haha, missed that completely.

    could you help me with 4a?


    y = x^5/2 ln 4x

    when Y = 0 find the coordinate, only one mark but unsure what I need to do
    A curve crosses the x-axis when y=0, so in this case: x^{\frac{5}{2}} \ln 4x = 0 this means either x^{5/2} = 0 or \ln 4x = 0, but the first one gives x=0 but the question says x&gt;0 so it's not a valid solution.

    So you need to solve \ln 4x = 0. What does that give you?

    Hint: \ln 4x = 0 \Rightarrow 4x = e^{0} = 1 \Rightarrow \cdots
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.