C3-trig

Announcements Posted on
Four things that unis think matter more than league tables 08-12-2016
    • Thread Starter
    Offline

    2
    ReputationRep:
    hi can someone solve this
    3secsquaredX-4=0

    can someone help me
    Offline

    2
    ReputationRep:
    (Original post by otrivine)
    hi can someone solve this
    3secsquaredX-4=0

    can someone help me
    You could write sec in terms of cos
    Offline

    1
    ReputationRep:
    (Original post by otrivine)
    hi can someone solve this
    3secsquaredX-4=0

    can someone help me
     \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4

    Remember,  \displaystyle sec^2x = \frac1{cos^2x}
    Offline

    0
    ReputationRep:
    3/cos^2 (x) = 4
    cos^2(x) = 3/4
    cos (x) = root3/2
    x = pi/6 or 30 degrees
    Offline

    1
    ReputationRep:
    (Original post by r_t)
    3/cos^2 (x) = 4
    cos^2(x) = 3/4
    cos (x) = root3/2
    x = pi/6 or 30 degrees
    It will be better to give OP some hints, so that he does it himself, rather than post a full solution for him.

    By the way, your solution isn't completely correct.
     \displaystyle cos^2x=\frac34 \implies cosx= \pm \frac{\sqrt3}{2}

    You haven't considered the negative case.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by raheem94)
     \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4

    Remember,  \displaystyle sec^2x = \frac1{cos^2x}
    i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
    Offline

    2
    ReputationRep:
    (Original post by otrivine)
    i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
    secX does not equal 3/4
    Offline

    1
    ReputationRep:
    (Original post by otrivine)
    i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
    I am not understanding what you are trying to do.

      \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4 \implies sec^2x=\frac43 \implies \frac1{cos^2x} = \frac43 \\ \implies cos^2x = \frac34

    Do you get it?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by raheem94)
    I am not understanding what you are trying to do.

      \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4 \implies sec^2x=\frac43 \implies \frac1{cos^2x} = \frac43 \\ \implies cos^2x = \frac34

    Do you get it?
    ohhh yes yes i get it now
    and one more thing for this question tansquared2X-2sec2x+1=0
    i got 90 degrees and one maths error is this correct so far or not?
    as for the equation i got secsquared2x-2sec2x+1=0
    Offline

    1
    ReputationRep:
    (Original post by otrivine)
    ohhh yes yes i get it now
    and one more thing for this question tansquared2X-2sec2x+1=0
    i got 90 degrees and one maths error is this correct so far or not?
    as for the equation i got secsquared2x-2sec2x+1=0
    It isn't correct.

     tan^22x-2sec2x+1=0

    Remember,  tan^22x=sec^22x-1

    So the equation becomes,
     tan^22x-2sec2x+1=0 \implies sec^22x-1-2sec2x+1=0 \\ \implies sec^22x-2sec2x=0
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by raheem94)
    It isn't correct.

     tan^22x-2sec2x+1=0

    Remember,  tan^22x=sec^22x-1

    So the equation becomes,
     tan^22x-2sec2x+1=0 \implies sec^22x-1-2sec2x+1=0 \\ \implies sec^22x-2sec2x=0
    yes that is what i got but my answer are wrong

    i applied quadratic formula X2-2X+0 which gives 2 and 0
    Offline

    2
    ReputationRep:
    (Original post by otrivine)
    yes that is what i got but my answer are wrong

    i applied quadratic formula X2-2X+0 which gives 2 and 0
    Factorise
    Offline

    1
    ReputationRep:
    (Original post by otrivine)
    yes that is what i got but my answer are wrong

    i applied quadratic formula X2-2X+0 which gives 2 and 0
    You don't need to apply the quadratic formula,

     \displaystyle sec^22x-2sec2x=0  \implies sec2x(sec2x-2)=0

    So you get 2 equations,  sec2x=0 \text{ and } sec2x-2 =0
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by steve2005)
    Factorise
    yes i get same answer 0 , 2
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by raheem94)
    You don't need to apply the quadratic formula,

     \displaystyle sec^22x-2sec2x=0  \implies sec2x(sec2x-2)=0

    So you get 2 equations,  sec2x=0 \text{ and } sec2x-2 =0
    i get 90 degrees and maths error

    and then i takeaway 90 with 360 then keep dividing by 2 correct?

    cause my book i giving me only 2 solution 0,180
    Offline

    1
    ReputationRep:
    (Original post by otrivine)
    yes i get same answer 0 , 2
    See my previous post.

    0 and 2 are correct because the 2 equation are sec2x=0 and sec2x=2.

    But you need to use  \displaystyle sec2x= \frac1{cos2x} to find the values of '2x' and hence 'x'
    Offline

    1
    ReputationRep:
    (Original post by otrivine)
    i get 90 degrees and maths error

    and then i takeaway 90 with 360 then keep dividing by 2 correct?

    cause my book i giving me only 2 solution 0,180
    Are the answers in your book 0 and 180?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by raheem94)
    See my previous post.

    0 and 2 are correct because the 2 equation are sec2x=0 and sec2x=2.

    But you need to use  \displaystyle sec2x= \frac1{cos2x} to find the values of '2x' and hence 'x'
    ok so dont we have to do firstly cos-1(0) then divide by 2 ?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by raheem94)
    Are the answers in your book 0 and 180?
    yes its the edexcel c3 text book
    Offline

    2
    ReputationRep:
    (Original post by otrivine)
    i get 90 degrees and maths error

    and then i takeaway 90 with 360 then keep dividing by 2 correct?

    cause my book i giving me only 2 solution 0,180
    I think you are looking at the wrong answers.
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: May 3, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Do you think you'll achieve your predicted A Level grades?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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

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